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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long double PI = 3.14159265358979323846;
const int MOD = 1e+9 + 7;
const int INFi = INT_MAX;
const int N = 2e+2 + 8;
vector<int> adj[N];
int vis[N];
int dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
int test_cases = 1;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
;
int n, ns;
string second;
int pre[N];
int dp[205][205][205][2];
deque<char> br;
int solve(int i, int j, int k, int ok) {
  ok |= (k == ns);
  if (i > 2 * n) return (ok == 1 && j == 0);
  if (j < 0) return 0;
  if (dp[i][j][k][ok] != -1) return dp[i][j][k][ok];
  int ans = 0;
  ans = (ans + solve(i + 1, j + 1,
                     (second[k + 1] == '(' && k < ns) ? k + 1 : pre[k], ok)) %
        MOD;
  ans = (ans + solve(i + 1, j - 1,
                     (second[k + 1] == ')' && k < ns) ? k + 1 : pre[k], ok)) %
        MOD;
  return dp[i][j][k][ok] = ans;
}
int check(int i, string t) {
  int tsz = i + 1;
  for (int k = (i + 1); k >= 1; k--) {
    if (second.substr(1, k) == t.substr(tsz - k + 1, k)) return k;
  }
  return 0;
}
void MAIN(int tc) {
  memset(dp, -1, sizeof(dp));
  cin >> n;
  cin >> second;
  ns = second.size();
  second = "#" + second + "#";
  ;
  string temp;
  for (int i = (0); i <= ns - 1; i++) {
    temp += second[i];
    if (second[i + 1] == '(')
      pre[i] = check(i, temp + ')');
    else
      pre[i] = check(i, temp + '(');
  }
  cout << solve(1, 0, 0, 0) << " ";
  cout << "\n";
  ;
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = (1); i <= test_cases; i++) {
    MAIN(i);
  }
}
