CodeForces 629C Famil Door and Brackets

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 lengthn 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 lengthn. He is going to pick some strings p and q consisting of round brackets and merge them in a stringp + s + q, that is add the stringp at the beginning of the string s and string q at the end of the strings.

Now he wonders, how many pairs of stringsp and q exists, such that the stringp + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo10⁹ + 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 lengthm consisting of characters '(' and ')' only.

Output

Print the number of pairs of string p andq such that p + s + q is a valid sequence of round brackets modulo10⁹ + 7.

Examples

Input

4 1(

Output

4

Input

4 4(())

Output

1

Input

4 3(((

Output

0

Note

In the first sample there are four different valid pairs:

1. p = "(",q = "))"
2. p = "()",q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"

In the second sample the only way to obtain a desired string is choose empty p and q.

In the third sample there is no way to get a valid sequence of brackets.

分析：先对给定的串做括号匹配，然后求出剩下有l 个 ） 和 r 个 （,dp[i][j]表示长度为i的有效匹配序列中多出j个左括号的方案数，那么我们只要枚举p就能同时的到q.

#include <cstdio>#include <iostream>#define MAXN 1000000007using namespace std;int n,m,totl,totr;char c;long long dp[2005][2005],ans;int main(){scanf("%d %d",&n,&m);scanf("%c",&c);for(int i = 1;i <= m;i++) {scanf("%c",&c);if(c == '(') totl++;else if(totl) totl--;     else totr++;}dp[0][0] = 1;for(int i = 1;i <= 2000;i++){dp[i][0] = dp[i-1][1];for(int j = 1;j <= i;j++)  dp[i][j] = (dp[i-1][j-1] + dp[i-1][j+1]) % MAXN; }for(int i = 0;i <= n - m;i++) for(int j = totr;j <= i;j++)  if(j - totr + totl <= n - m - i) ans = (ans + dp[i][j]*dp[n - m - i][j - totr + totl]) % MAXN;cout<<ans<<endl;}