Regular Bracket Sequence

A. Regular Bracket Sequence

A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".

You are given a sequence s of n characters "(", ")", and/or "?". There is exactly one character ( and exactly one character ) in this sequence.

You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.

Determine if it is possible to obtain an "RBS" after these replacements.

Input

The first line contains one integer t ((1leq tleq 1000)) — the number of test cases.

Each test case consists of one line containing s ((2leq |s|leq 100)) a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output

For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise.

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

input

Copy

5
()
(?)
(??)
??()
)?(?

output

Copy

YES
NO
YES
YES
NO

Note

In the first test case, the sequence is already an RBS.

In the third test case, you can obtain an RBS as follows: ()() or (()).

In the fourth test case, you can obtain an RBS as follows: ()().

没有引号,看不懂加粗字体,原来是只有一个左括号和一个右括号

考虑前缀和,左括号+1,右括号-1,则除了?,一个合法的序列是前缀和均大于零且sum[n]=0

把?替换成1或-1,因为1和-1的个数是确定的,通过sum[n]可求出来

贪心,先让所有1放左边,-1接着放-1,这样就能让前缀和尽可能多的大于等于0

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN=108,mod=1e9+7,inf=0x3f3f3f3f;
char s[1000];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        int n=strlen(s),ans=1;
        int a=0,b=0,c=0;
        for(int i=0;i<n;++i){
            if(s[i]=='(')a++;
            if(s[i]=='?')b++;
            if(s[i]==')')c++;
        }
        if((a-c+b)&1)ans=0;
        int x=(a-c+b)/2,y=b-x;
        if(y<0)ans=0;
        if(x<0)ans=0;
        for(int i=0;i<n;++i){
            if(s[i]=='?'){
                if(x)s[i]='(',x--;
                else s[i]=')',y--;
            }
        }
        for(int i=0,tot=0;i<n;++i){
            if(s[i]=='(')tot++;
            if(s[i]==')')tot--;
            if(tot<0)ans=0;
        }
        if(ans)puts("YES");
        else puts("NO");
    }
    return 0;
}