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
import sys
from functools import lru_cache

sys.setrecursionlimit(10000)

mn = input()
#print(mn)
m,n = [int(i) for i in mn.split()]

my_string = input()

leftfill = 0
c = 0
const = 1000000007


for char in my_string:
    if char == ')':
        c-=1
    elif char == '(':
        c+=1
    leftfill = min(leftfill,c)
    
rightfill = max(c,0)
leftfill *= -1


@lru_cache(maxsize=None)
def gendp(nsteps,balance,maxdrawdown):
    if balance < maxdrawdown:
        return 0
    if abs(balance) == nsteps:
        return 1
    if nsteps == 0:
        return 0
    if (nsteps + balance) % 2 ==1:
        return 0
    if balance == maxdrawdown:
        return gendp(nsteps-1,balance+1,maxdrawdown)
    if nsteps >= balance and (nsteps + balance) % 2 == 0:
        return gendp(nsteps-1,balance-1,maxdrawdown) + gendp(nsteps-1, balance+1,maxdrawdown)
    return 0


diff = m-n-abs(leftfill) - rightfill
sum_ = 0

for i in range(diff + 1):
    ddd = min(diff-i,i)
    for j in range(ddd + 1):
        k1p1 = leftfill + i
        k1p2 = leftfill + j
        k2p1 = rightfill + diff - i
        k2p2 = -rightfill - j
        if k1p1 ==0 and k1p2 ==0:
            k1 = 1
        else:
            k1 = gendp(k1p1,k1p2,0)
        if k2p1 == 0 and k2p2 == 0:
            k2 = 1
        else:
            k2 = gendp(k2p1, k2p2,k2p2) 
        #k1 = max(k1,1)
        #k2 = max(k2,1)
        sum_ +=  k1 * k2  % const
        sum_ = sum_ % const

print(sum_)
