CF629C.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 $10^{9}+7$ .

First line contains $n$ and $m$ ( $1<=m<=n<=100000,n-m<=2000$ ) — 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.

Print the number of pairs of string $p$ and $q$ such that $p+s+q$ is a valid sequence of round brackets modulo $10^{9}+7$ .

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.