C. Famil Door and Brackets

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

思路：dp；

dp[i][j]表示前i个括号，开口向右减开口向左的值SS。方程：dp[i][j]=dp[i-1][j-1]+dp[i-1][j+1].在i的时候当前选向左开口或向右开口。

可以知道dp[0][0]=1；由于n-m<=2000;所以枚举p，因为最后要平衡所以q也可以知道了，并且q的取值也就是前面的dp，因为要和前面匹配，前面p+s结束时必定有向右的

括号数大于向左的括号数，所以在剩下的q只要取向左的括号数数-向右的括号数，与前面的匹配中和就行了，也就是将前面的dp看成在i时开口向右减开口向左的值SS，所以套用前面的dp就行了。  最后sum=((sum)%N+(dp[i][j]*dp[cc][ans])%N)%N;

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 using namespace std;
8 typedef long long LL;
9 LL dp[2205][2205];
10 char str[100005];
11 int QZ[100005];
12 const LL N=1e9+7;
13 int main(void)
14 {
15 LL i,j,k,p,q;
16 dp[0][0]=1;
17 for(i=1;i<=2200;i++) 18 { 19 for(j=0;j<=i;j++) 20 { 21 if(j>0)
22 {
23 dp[i][j]=(dp[i][j]+dp[i-1][j-1])%N;
24 }
25 dp[i][j]=(dp[i][j]+dp[i-1][j+1])%N;
26 }
27 }
28 while(scanf("%I64d %I64d",&p,&q)!=EOF)
29 {memset(QZ,0,sizeof(QZ));
30 scanf("%s",str);LL meq=1e8;LL maxx;
31 for(i=0;i=0&&ans<=cc)
54 {
55 sum=((sum)%N+(dp[i][j]*dp[cc][ans])%N)%N;
56 }
57 }
58 }
59 printf("%I64d\n",sum);
60 }
61 return 0;
62 }