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;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
void print(vector<long long> &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld ", v[i]);
  }
  printf("\n");
}
void print(vector<pair<long long, long long> > &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld %lld\n", v[i].first, v[i].second);
  }
}
void print(double d) { cout << fixed << setprecision(10) << d << endl; }
void print(string s, double d) {
  cout << s << " ";
  cout << fixed << setprecision(10) << d << endl;
}
long long mn = 8000000000000000000;
long long sum = 0;
string s;
long long dp[220][220][220][2];
long long n;
long long dp1[220][2];
long long get(long long len, long long left) {
  string s1 = "";
  if (len >= s.size()) return 0;
  long long &ret = dp1[len][left];
  if (ret != -1) return ret;
  for (int i = 0; i < len; i++) {
    s1.push_back(s[i]);
  }
  if (!left)
    s1.push_back('(');
  else
    s1.push_back(')');
  int mx = 0;
  for (int i = 0; i < s1.size(); i++) {
    int cnt = 0;
    for (int j = 0; j < s.size(); j++) {
      if (s1[i + cnt] != s[j]) break;
      cnt++;
    }
    if (i + cnt == s1.size()) mx = max(mx, cnt);
  }
  return ret = mx;
}
long long calc(long long i, long long cnt, long long nisi, long long ok) {
  long long ok1 = ok;
  if (nisi == s.size()) ok1 = 1;
  if (i == 2 * n) {
    if (cnt == 0 and ok1) return 1LL;
    return 0;
  }
  if (cnt < 0) return 0;
  long long &ret = dp[i][cnt][nisi][ok];
  if (ret != -1) return ret;
  long long x = 0;
  long long open = get(nisi, 0);
  long long close = get(nisi, 1);
  x = calc(i + 1, cnt + 1, open, ok1);
  x += calc(i + 1, cnt - 1, close, ok1);
  x %= 1000000007;
  return ret = x;
}
int main() {
  long long q, i, j = 0, temp, t, k, ans = 0, x, y, z, cnt = 0, m, fg = 0,
                  mx1 = 0, mn1 = 8000000000000000000;
  cin >> n >> s;
  for (char ch : s) {
    if (ch == '(')
      sum++;
    else
      sum--;
    mn = min(mn, sum);
  }
  memset((dp), -1, sizeof(dp));
  memset((dp1), -1, sizeof(dp1));
  ans = calc(0, 0, 0, 0);
  cout << ans << endl;
}
