Codeforces Round #272 (Div. 2)

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链接 : http://codeforces.com/contest/476

D题yy，ABC水

Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integerm.

What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?

Input

The single line contains two space separated integersn, m (0 < n ≤ 10000, 1 < m ≤ 10).

Output

Print a single integer — the minimal number of moves being a multiple ofm. If there is no way he can climb satisfying condition print - 1 instead.

Sample test(s)

Input

10 2

Output

Input

3 5

Output

-1

Note

For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.

For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.

#include <cstdio>
int main()
{int n, m;scanf("%d %d", &n, &m);int temp = n / 2;if(n % 2) temp++;if(temp % m == 0) printf("%d\n", temp);else{temp = temp + m - temp % m;if(temp > n) printf("-1\n");else printf("%d\n", temp);}
}

Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.

Each command is one of the following two types:

Go 1 unit towards the positive direction, denoted as'+'
Go 1 unit towards the negative direction, denoted as'-'

But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the1 unit to the negative or positive direction with the same probability0.5).

You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?

Input

The first line contains a string s₁ — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+','-'}.

The second line contains a string s₂ — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+','-', '?'}.'?' denotes an unrecognized command.

Lengths of two strings are equal and do not exceed10.

Output

Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed10^- 9.

Sample test(s)

Input

++-+-
+-+-+

Output

1.000000000000

Input

+-+-
+-??

Output

0.500000000000

Input

+++
??-

Output

0.000000000000

Note

For the first sample, both s₁ and s₂ will lead Dreamoon to finish at the same position + 1.

For the second sample, s₁ will lead Dreamoon to finish at position 0, while there are four possibilites fors₂: {"+-++","+-+-", "+--+","+---"} with ending position {+2, 0, 0, -2} respectively. So there are2 correct cases out of 4, so the probability of finishing at the correct position is0.5.

For the third sample, s₂ could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0.

#include <cmath>
#include <cstdio>
#include <cstring>
char s1[15],s2[15];
int main()
{scanf("%s %s",s1,s2);int p1=0, m1=0;int p2=0, m2=0;int size=(int)strlen(s1);int wh=0;for(int i=0;i<size; i++){if(s1[i]=='+') ++p1;else ++m1;}for(int i=0;i<size; i++){if(s2[i]=='+') ++p2;else if(s2[i]=='-') ++m2;else ++wh;}if(wh == 0){if(p1 - m1 == p2 - m2)printf("1\n");elseprintf("0\n");return 0;}int all = 1;for(int i = 0;i < wh; i++)all *= 2;int pneed,mneed;int temp = p1 - m1 - p2 + m2;if(fabs(temp) > wh){printf("0\n");return 0;}pneed = (temp + wh) / 2;mneed = wh - pneed;int ans = 1;for(int i = 0; i < pneed; i++){ans *= wh;--wh;}for(int i=1; i<= pneed; i++)ans /= i;printf("%.12f\n",double(ans)/double(all));
}

Dreamoon loves summing up something for no reason. One day he obtains two integersa and b occasionally. He wants to calculate the sum of allnice integers. Positive integer x is called nice if $\text{[math]}$ and $\text{[math]}$ , wherek is some integer number in range [1, a].

By $\text{[math]}$ we denote thequotient of integer division of x and y. By $\text{[math]}$ we denote theremainder of integer division of x and y. You can read more about these operations here:http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo1 000 000 007 (10⁹ + 7). Can you compute it faster than Dreamoon?

Input

The single line of the input contains two integersa, b (1 ≤ a, b ≤ 10⁷).

Output

Print a single integer representing the answer modulo1 000 000 007 (10⁹ + 7).

Sample test(s)

Input

1 1

Output

Input

2 2

Output

Note

For the first sample, there are no nice integers because $\text{[math]}$ is always zero.

For the second sample, the set of nice integers is{3, 5}.

#include <cstdio>
#define ll long long
const int mod=1e9 + 7;
int main()
{ll a,b;scanf("%I64d %I64d", &a, &b);if(b == 1) printf("0\n");else{ll res = (b * (b - 1) / 2) % mod;ll sum = (a * (a + 1) / 2) % mod;ll ans = (a % mod * res % mod) % mod;ll temp = (sum % mod * res % mod) % mod;printf("%I64d\n", (ans % mod + (temp % mod * b % mod)) % mod);}
}

Dreamoon likes to play with sets, integers and $\text{[math]}$ . $\text{[math]}$ is defined as the largest positive integer that divides botha and b.

Let S be a set of exactly four distinct integers greater than0. Define S to be of rankk if and only if for all pairs of distinct elementss_i,s_j fromS, $\text{[math]}$ .

Given k andn, Dreamoon wants to make up n sets of rank k using integers from1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimumm that makes it possible and print one possible solution.

Input

The single line of the input contains two space separated integersn, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

Output

On the first line print a single integer — the minimal possiblem.

On each of the next n lines print four space separated integers representing thei-th set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimalm, print any one of them.

Sample test(s)

Input

1 1

Output

5
1 2 3 5

Input

2 2

Output

22
2 4 6 22
14 18 10 16

Note

For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since $\text{[math]}$ .

#include <cstdio>
int main()
{int n, k;scanf("%d %d", &n, &k);int a = 1, b = 2, c = 3, d = 5;printf("%d\n", (d * k + 6 * k * (n- 1)));a *= k; b *= k; c *= k; d *= k;for(int i = 0; i < n; i++){printf("%d %d %d %d\n",a, b, c, d);a += 6 * k;b += 6 * k;c += 6 * k;d += 6 * k;}
}

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