As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length *n* more than any other strings!

The sequence of round brackets is called valid if and only if: 
 1.  the total number of opening brackets is equal to the total number of closing brackets; 1.  for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets. 
Gabi bought a string *s* of length *m* (*m* ≤ *n*) and want to complete it to obtain a valid sequence of brackets of length *n*. He is going to pick some strings *p* and *q* consisting of round brackets and merge them in a string *p* + *s* + *q*, that is add the string *p* at the beginning of the string *s* and string *q* at the end of the string *s*.

Now he wonders, how many pairs of strings *p* and *q* exists, such that the string *p* + *s* + *q* is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 10<sup class="upper-index">9</sup> + 7.
First line contains *n* and *m* (1 ≤ *m* ≤ *n* ≤ 100 000, *n* - *m* ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.

The second line contains string *s* of length *m* consisting of characters '(' and ')' only.
In the first sample there are four different valid pairs: 
 1.  *p* = "(", *q* = "))" 1.  *p* = "()", *q* = ")" 1.  *p* = "", *q* = "())" 1.  *p* = "", *q* = ")()" 
In the second sample the only way to obtain a desired string is choose empty *p* and *q*.

In the third sample there is no way to get a valid sequence of brackets.
4 1
(
4
4 4
(())
1
4 3
(((
0
875 50
)))((())()))((()(())))))())))((((((()))))))()(((((
0
1980 464
))(()()))(((((((((()))))))(()(((()((()))()()())()))()))(()))))))(())((())))()())()((())())()())))(())()(()))(()())()((((()))())()(())))))(()()(()(((((()(()()))(((()))(())))))()())(())))))())()()((())))))))((()(())))))()()(()((()((()()))(()))(())(()))()((((())()()))))))()(())))()(()())()())(((((()))())))())())(()))()(()))())((())((((()(()(())))(((()()))))()()()))))((()())()((())())))())))()(()(()()(((((()((((()))()(())()))))()(()))(()(((((((()((()(())))(((((())
854368836
1542 282
())())()((()(()))()((())()))((())(()))))(()()))(())((()))()((()())())()))((())(((()(())((()()())((((())))((()((((()(()()))))(()(()()(())()((())())())))))()()())))(()((((()))(()(()(()(()))())((()()()()(()(()))())(((()(())()(())()()())))()))())(()))(((())()))((())()(())))))(())))()()
631927032
2 2
)(
0
2 2
))
0
from _collections import defaultdict
n, m = input().split()
n, m = int(n), int(m)

dp = [None] * (n-m+1)
for _ in range(n-m+1):
    dp[_] = defaultdict(int)

dp[0][0] = 1

for i in range(n-m):
    for k, v in dp[i].items():
        if k > 0:
            dp[i+1][k-1] += dp[i][k]
        dp[i+1][k+1] += dp[i][k]

balance = 0
min_interm = 0

for ch in input():
    if ch == "(":
        balance += 1
    else:
        balance -= 1
        min_interm = min(balance, min_interm)

ans = 0
for spl in range(0, n-m+1):
    for bal, comb in dp[spl].items():
        if bal < -min_interm:
            continue
        ans += comb * dp[n-m-spl][bal+balance]
print(ans)
        