You are given a bracket sequence s (not necessarily a regular one). A bracket sequence is a string containing only characters '(' and ')'.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

Your problem is to calculate the number of regular bracket sequences of length 2n containing the given bracket sequence s as a substring (consecutive sequence of characters) modulo 10^9+7 (1000000007).

Input

The first line of the input contains one integer n (1 ≤ n ≤ 100) — the half-length of the resulting regular bracket sequences (the resulting sequences must have length equal to 2n).

The second line of the input contains one string s (1 ≤ |s| ≤ 200) — the string s that should be a substring in each of the resulting regular bracket sequences (|s| is the length of s).

Output

Print only one integer — the number of regular bracket sequences containing the given bracket sequence s as a substring. Since this number can be huge, print it modulo 10^9+7 (1000000007).

Examples

Input

5
()))()


Output

5


Input

3
(()


Output

4


Input

2
(((


Output

0

Note

All regular bracket sequences satisfying the conditions above for the first example: 

  * "(((()))())"; 
  * "((()()))()"; 
  * "((()))()()"; 
  * "(()(()))()"; 
  * "()((()))()". 



All regular bracket sequences satisfying the conditions above for the second example: 

  * "((()))"; 
  * "(()())"; 
  * "(())()"; 
  * "()(())". 



And there is no regular bracket sequences of length 4 containing "(((" as a substring in the third example.
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e3 + 36;
const long long M = 2e2 + 36;
const long long INF = 2e9 + 7;
const long long MOD = 987654323;
const long long MOD1 = 1e9 + 7;
const long long MOD2 = 1e9 + 9;
const int XK[4] = {-1, 0, 1, 0};
const int YK[4] = {0, -1, 0, 1};
const int X[8] = {1, 1, -1, -1, 2, 2, -2, -2};
const int Y[8] = {2, -2, 2, -2, 1, -1, 1, -1};
long long dp[M][M][M][2];
long long len[M][2];
string s;
int solve(string t, int first) {
  if (first == 1) {
    t += ')';
  } else {
    t += '(';
  }
  for (int i((int)t.size()); i > 0; --i) {
    if (s.substr(0, i) == t.substr((int)t.size() - i, i)) {
      return i;
    }
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  cin >> s;
  if (s[0] == '(') {
    len[0][0] = 1;
  } else {
    len[0][1] = 1;
  }
  for (int i(1); i <= (int)s.size(); ++i) {
    string t = s.substr(0, i);
    len[i][0] = solve(t, 0);
    len[i][1] = solve(t, 1);
  }
  dp[0][0][0][0] = 1;
  for (int i(0); i < n * 2; ++i) {
    for (int j(0); j <= n; ++j) {
      for (int l(0); l <= (int)s.size(); ++l) {
        for (int f(0); f < 2; ++f) {
          if (dp[i][j][l][f] == 0) continue;
          if (j < n) {
            dp[i + 1][j + 1][len[l][0]]
              [f | (len[l][0] == (int)s.size() ? 1 : 0)] += dp[i][j][l][f];
            dp[i + 1][j + 1][len[l][0]]
              [f | (len[l][0] == (int)s.size() ? 1 : 0)] %= MOD1;
          }
          if (j > 0) {
            dp[i + 1][j - 1][len[l][1]]
              [f | (len[l][1] == (int)s.size() ? 1 : 0)] += dp[i][j][l][f];
            dp[i + 1][j - 1][len[l][1]]
              [f | (len[l][1] == (int)s.size() ? 1 : 0)] %= MOD1;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i(0); i <= (int)s.size(); ++i) {
    ans += dp[n << 1][0][i][1];
    ans %= MOD1;
  }
  cout << ans << endl;
  return 0;
}
