The Monster（Codeforce-C-思维题）

C. The Monster

time limit per test

1 second

memory limit per test

256 megabytes

 

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

((?))

output

4

input

??()??

output

7

Note

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

1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".

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

1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".

题意：给你一个由'('、')'和'?'组成的字符串s，问你s有多少个子串是好串（可以通过将'?'转化成'('或者')';  好串的定义：形如()、(())；

分析：思维题，每次都进行一下判断： l代表左括号数目，r代表问号数目，ans代表符合条件的串个数

① 若s[i]=='('  l++;

② 若s[i]==')'  l--;

③ 若s[i]=='?'  r++,l--;   （先将'?'看成右括号，同时记录'?'的数目）

④ 如果l==0  匹配成功，ans++;

⑤ 如果l<0&&r>0  l+=2,r--;  (右括号多于左括号，同时又有问号，说明右括号多于左括号的有部分原因是？造成的，所以我们把？变成左括号）

⑥ 如果l<0&&r==0  结束循环  (右括号多于左括号,不是因为？的原因，无法再形成好串)

AC代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 5010
using namespace std;
char s[N];
int main()
{
    cin>>s;
    int len=strlen(s);
    int ans=;
    for (int i=;i<len;i++)
      {
          int l=,r=;
          for (int j=i;j<len;j++)
            {
                if (s[j]=='(') l++;
                else if (s[j]==')') l--;
                else if (s[j]=='?') l--,r++;
                if (!l)  ans++;
                else if(l<&&r>) l+=,r--;
                else if (l<&&!r) break;
            }
      }
      cout << ans << endl;
    return ;
}