codeforces629C Famil Door and Brackets (dp)

As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!

The sequence of round brackets is called valid if and only if:

1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.

Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.

Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.

Input

First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.

The second line contains string s of length m consisting of characters '(' and ')' only.

Output

Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.

Examples

input

4 1
(

output

4

input

4 4
(())

output

1

input

4 3
(((

output

0

题意：给你一个长度为n的括号匹配串(不一定恰好匹配)，让你在这个串的前面和后面加上一些括号匹配串，使得这个括号串平衡(平衡的含义是对于任意位置的括号前缀和大于等于0，且最后的前缀和为0)。

思路：比较容易想到的思路是枚举这个字符串前面p字符串的长度，那么后面q字符串的长度就知道了。那么p字符串要满足什么条件呢，因为要使得任意位置的前缀和大于等于0，所以我们可以使得p字符串的前缀和大于等于字符串s的最小前缀和minx，那么p+s就符合前缀和大于等于0，然后q的方案数也能确定了。我们用dp[i][j]表示i个括号平衡度为j的方案数，那么可以先预处理出来dp的值。然后我们算出s字符串的最小前缀和minx，最后我们只要枚举p的长度和平衡度c，那么sum+=dp[p][c]*dp[n-m-p][now+c],(now是整个s字符串的平衡度，考虑q的方案数时，我们要考虑对称性)。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define eps 1e-15
#define maxn 100050
#define MOD 1000000007
char s[maxn];
ll dp[2050][2060];

int main()
{
    int n,m,i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        scanf("%s",s+1);
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(i=1;i<=n-m;i++){
            for(j=0;j<=i;j++){
                if(j>0){
                    dp[i][j]=(dp[i][j]+dp[i-1][j-1])%MOD;
                }
                dp[i][j]=(dp[i][j]+dp[i-1][j+1])%MOD;
                if(dp[i][j]>=MOD)dp[i][j]-=MOD;
            }
        }
        int minx=inf;
        int now=0;
        for(i=1;i<=m;i++){
            if(s[i]=='(')now++;
            else now--;
            minx=min(minx,now);
        }

        ll sum=0;
        for(i=0;i<=n-m;i++){
            for(j=0;j<=i;j++){
                if(j>=-minx && j+now<=n-m-i){
                    sum=(sum+dp[i][j]*dp[n-m-i][j+now ])%MOD;
                }
            }
        }
        printf("%I64d
",sum);
    }
    return 0;
}