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
def fact(n):
    d = 1
    for i in range(2, n + 1):
        d *= i
    return d

def f():
    n, m = [int(i) for i in input().split()]
    s = input()
    p = [(0, 0)] * m
    c1 = 0
    c2 = 0
    for i in range(m):
        if s[i] == '(':
            c1 += 1
        else:
            c2 += 1
        p[i] = (c1, c2)
    if c1 > n // 2 or c2 > n // 2:
        print(0)
    else:
        q = 0
        c1 = (n // 2) - c1
        c2 = (n // 2) - c2
        t = abs(c1 - c2)
        c1 = min(c1, c2)
        c2 = c1
        f1 = fact(t + 1 + c1)
        l1 = t + 1 + c1
        f2 = fact(c1)
        l2 = c1
        f3 = fact(t + 1)
        l3 = t + 1
        cc = (f1 // (f2 * f3)) % 1000000007
        for i in range(c1 + 1):
            q = (q + cc) % 1000000007
            if i != c1:
                cc *= (l1 + 1) * l2
                l1 += 1
                l2 -= 1
                cc //= (l3 + 1) * (l3 + 2)
                l3 += 2
        print(q)


f()