629_C. Famil Door and Brackets
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; 
  2. 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 109 + 7.

Input

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.

Output

Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.

Examples

Input

4 1
(


Output

4


Input

4 4
(())


Output

1


Input

4 3
(((


Output

0

Note

In the first sample there are four different valid pairs: 

  1. p = "(", q = "))" 
  2. p = "()", q = ")" 
  3. p = "", q = "())" 
  4. 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.
```python
n, m = map(int, input().split())
s = input()
mod = 10 ** 9 + 7
c, b, ans, d, k = 0, 0, 0, [[1]], n - m
for i in s:
    c += (i == '(') * 2 - 1
    b = min(c, b)
for i in range(n - m):
    nd = d[-1][1:] + [0] * 2
    for j in range(1, i + 2):
        nd[j] = (nd[j] + d[-1][j - 1]) % mod
    d.append(nd)
for i in range(k + 1):
    for j in range(-b, min(k - i - c, i) + 1):
        ans = (ans + d[i][j] * d[k - i][j + c]) % mod
print(ans)

```