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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 205;
int n;
int len;
string s;
int p[N][2];
int dp[N][N][N][2];
int calc(int id, int cp, int bal, int found) {
  if (bal < 0) {
    return 0;
  }
  if (found == 0 && cp == len) {
    return calc(id, cp, bal, 1);
  }
  if (id == 2 * n) {
    if (found && bal == 0) {
      return 1;
    } else {
      return 0;
    }
  }
  int& ans = dp[id][cp][bal][found];
  if (ans != -1) {
    return ans;
  }
  ans = 0;
  if (found) {
    ans += calc(id + 1, cp, bal + 1, 1);
    ans += calc(id + 1, cp, bal - 1, 1);
    ans %= M;
    return ans;
  }
  ans += calc(id + 1, p[cp][0], bal + 1, 0);
  ans += calc(id + 1, p[cp][1], bal - 1, 0);
  ans %= M;
  return ans;
}
void solve() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> s;
  p[0][0] = p[0][1] = 0;
  if (s[0] == '(')
    p[0][0] = 1;
  else
    p[0][1] = 1;
  len = (int)((s).size());
  for (int i = 1; i < len; i++) {
    for (int j = i + 1; j > 0; j--) {
      bool ok = 1;
      for (int k = 0; k < j - 1; k++) {
        if (s[k] != s[k + i - j + 1]) {
          ok = 0;
        }
      }
      if (ok && s[j - 1] == '(') {
        p[i][0] = j;
        break;
      }
    }
  }
  for (int i = 1; i < len; i++) {
    for (int j = i + 1; j > 0; j--) {
      bool ok = 1;
      for (int k = 0; k < j - 1; k++) {
        if (s[k] != s[k + i - j + 1]) {
          ok = 0;
        }
      }
      if (ok && s[j - 1] == ')') {
        p[i][1] = j;
        break;
      }
    }
  }
  cout << calc(0, 0, 0, 0);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
