CodeForces - 785D(92/600)

As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters “(” and “)” (without quotes)).

On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:

It is not empty (that is n ≠ 0).

The length of the sequence is even.

First charactes of the sequence are equal to “(“.

Last charactes of the sequence are equal to “)”.

For example, the sequence “((()))” is an RSBS but the sequences “((())” and “(()())” are not RSBS.

Elena Ivanovna, Anton’s teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.

Because the answer can be very big and Anton’s teacher doesn’t like big numbers, she asks Anton to find the answer modulo 109 + 7.

Anton thought of this task for a very long time, but he still doesn’t know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!

Input

The only line of the input contains a string s — the bracket sequence given in Anton’s homework. The string consists only of characters “(” and “)” (without quotes). It’s guaranteed that the string is not empty and its length doesn’t exceed 200 000.

Output

Output one number — the answer for the task modulo 109 + 7.

额》。。。

Example

Input

)(()()

Output

6

Input

()()()

Output

7

Input

)))

Output

0

这个题就是先枚举左端点，二分右端点

的出来一个组合数相乘的公式

然后再用范德蒙恒等式化简

组合数取模直接用逆元处理就完事了

#include<bits/stdc++.h>
using namespace std;
#define int long long
int zt[400001],yt[400001],zts=0,yts=0;
int jc[400001],mo=1e9+7;
long long inv(long long k){
if(k == 1) return 1;
return (mo - mo / k) * inv(mo % k) % mo;
}
int c(int xia,int shang)
{
int ccc=jc[xia]*inv(jc[shang])%mo;
ccc*=inv((jc[xia-shang]+mo)%mo)%mo;
return ccc%mo;
}
main()
{
//  freopen("test.txt","r",stdin);
string q;
cin>>q;
for(int a=0;a<q.size();a++)
{
if(q[a]=='(')zt[++zts]=a;
else yt[++yts]=a;
}
jc[0]=jc[1]=1;
for(int a=1;a<=200000;a++)
{
jc[a]=jc[a-1]*a%mo;
jc[a]=(jc[a]+mo)%mo;
}
//  for (int i = 2; i <= 200000; i++)inv[i] = (mo - mo / i) * inv[mo % i] % mo;
int fs=0;
for(int a=1;a<=zts;a++)
{
int z=1,y=yts;
//  cout<<"caonimacaonima"<<z;
//      return 0;
if(yt[yts]<zt[a])break;
int wz=-1;
while(z<=y)
{
int mid=(z+y)/2;
if(yt[mid]>=zt[a])
{
wz=mid;
y=mid-1;
}
else z=mid+1;
}
if(wz==-1)continue;
//      cout<<wz;return 0;
int x=yts-wz+1;
//      cout<<x;return 0;
int t=min(x,a);
//      cout<<t;return 0;
//      cout<<c(a+x,t);return 0;
fs=(fs+c(a+x-1,x-1))%mo;
//      cout<<fs<<endl;
//      cout<<fs;   return 0;
}
cout<<fs%mo;
}