cf918C(贪心)

贪心真的是弱。。。假期还得好好练练qaq

这个主要是在问号的处理，转念一想，问号如果能取）就尽量取），这样早点配完就可以把前面那部分给扔掉了。。如果不能取）就要把当前序列中最后一个？取（来和这个？或）匹配。。

当然还假设了？取（的配对情况，为了避免右括号过多无法匹配到左括号的情况。。

然后简介地说就是在枚举过程中令( + ?>=) 且 ) + ?>=(

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define eps 1e-8
#define inf 1000000007
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define ls T[i<<1]
#define rs T[i<<1|1]
#define op T[i]
#define mid (x+y>>1)
#define NM 10005
#define nm 10005
#define pi 3.141592653
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}





int n,x,y,ans;
char s[NM];
int main(){
    scanf("%s",s);
    n=strlen(s)-1;
    inc(i,0,n){
        x=y=0;
        inc(j,i,n){
            if(s[j]=='(')x++,y++;
            else if(s[j]==')')x--,y--;
            else x--,y++;
            if(x<0)x=1;
            if(y<0)break;
            if(!x)ans++;
        }
    }
    return 0*printf("%d\n",ans);
}

C. The Monster

time limit per test

memory limit per test

input

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.
• ifsis a correct bracket sequence, then(s)
• ifsandtare correct bracket sequences, thenst(concatenation ofsandt) 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

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 s_ls_l + 1...s_r is pretty, where s_i 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

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

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 "()()()".