CodeForces - 918C The Monster（合法的括号）

C. The Monster

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.

Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.

A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:

Empty string is a correct bracket sequence.
if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.

A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.

Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.

Joyce doesn't know anything about bracket sequences, so she asked for your help.

Input

The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).

Output

Print the answer to Will's puzzle in the first and only line of output.

Examples

input

Copy

((?))

output

Copy

input

Copy

??()??

output

Copy

Note

For the first sample testcase, the pretty substrings of s are:

"(?" which can be transformed to "()".
"?)" which can be transformed to "()".
"((?)" which can be transformed to "(())".
"(?))" which can be transformed to "(())".

For the second sample testcase, the pretty substrings of s are:

"??" which can be transformed to "()".
"()".
"??()" which can be transformed to "()()".
"?()?" which can be transformed to "(())".
"??" which can be transformed to "()".
"()??" which can be transformed to "()()".
"??()??" which can be transformed to "()()()".

从左到右记录左括号，右括号，问号的数量，先把问号看作右括号，如果左括号加问号都小于右括号，就退出；

如果不退出，但是左右括号数量不相等，并且存在问号，就把问号转成左括号；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <list>
using namespace std;
typedef long long ll;

int main()
{
    char s[5010];
    cin >> s;
    int l = strlen(s),i,j;
    int ans = 0;//合法的区间的数量；
    for(i = 0;i <l-1;i++){//开始的区间位置
        int x=0,y=0,z=0;//x左括号，y右括号，z问号
        for(j = i;j<l;j++){
            if(s[j] == '(') x++;
            else if(s[j] == ')') y++;
            else y++,z++;//先把问号看成右括号
            if(x+z<y-z) break;左括号+问号<右括号，不合法
            while(x<y && z){
                x++,y--,z--;//问号转成左括号
            }
            if(x==y) ans++;//合法的情况
        }
    }
    cout <<ans<<endl;
    return 0;
}

CodeForces - 918C The Monster（合法的括号）

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