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;
typedef long long lo;
typedef vector<lo> vl;
typedef vector<lo> vll;
typedef vector<vl> vvl;
template <typename T>
ostream &operator<<(ostream &o, vector<T> v) {
  if (v.size() > 0) o << v[0];
  for (unsigned i = 1; i < v.size(); i++) o << " " << v[i];
  return o << "\n";
}
template <typename U, typename V>
ostream &operator<<(ostream &o, pair<U, V> p) {
  return o << "(" << p.first << ", " << p.second << ") ";
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (unsigned i = 0; i < v.size(); i++) in >> v[i];
  return in;
}
template <typename T>
istream &operator>>(istream &in, pair<T, T> &p) {
  in >> p.first;
  in >> p.second;
  return in;
}
string s;
lo len[203][2];
lo dp[203][203][203][2];
lo cal(string &t) {
  for (lo i = t.size(); i > 0; i--) {
    if (s.substr(0, i) == t.substr(t.size() - i, i)) return i;
  }
  return 0;
}
void add(lo &a, lo b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  if (a < 0) a += 1000000007;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  lo n, ssz;
  cin >> n >> s;
  ssz = s.size();
  if (s[0] == '(')
    len[0][0] = 1;
  else
    len[0][1] = 1;
  string pref;
  for (lo i = (0); i < (lo)ssz; i++) {
    pref += s[i];
    pref += '(';
    len[i + 1][0] = cal(pref);
    pref.pop_back();
    pref += ')';
    len[i + 1][1] = cal(pref);
    pref.pop_back();
  }
  dp[0][0][0][0] = 1;
  for (lo i = (0); i < (lo)2 * n; i++) {
    for (lo(j) = 0; (j) < (n + 1); (j)++) {
      for (lo(pos) = 0; (pos) < (ssz + 1); (pos)++) {
        for (lo(f) = 0; (f) < (2); (f)++) {
          if (dp[i][j][pos][f] == 0) continue;
          if (j + 1 <= n)
            add(dp[i + 1][j + 1][len[pos][0]][f | (len[pos][0] == ssz)],
                dp[i][j][pos][f]);
          if (j > 0)
            add(dp[i + 1][j - 1][len[pos][1]][f | (len[pos][1] == ssz)],
                dp[i][j][pos][f]);
        }
      }
    }
  }
  lo ans = 0;
  for (lo i = (0); i < (lo)ssz + 1; i++) add(ans, dp[2 * n][0][i][1]);
  cout << ans << "\n";
  return 0;
}
