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You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a)if it is the empty string(b)if A and B are correct, AB is correct,(c)if A is correct, (A ) and [A ] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input

The file contains a positive integer n and a sequence of n strings of parentheses () and [], one string a line.

Output

A sequence of Yes or No on the output file.

Sample Input

3

([])
(([()])))
([()[]()])()

Sample Output

Yes
No
Yes

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Miguel Revilla
2000-08-14 题意： 如果有一个'（'那么就要有一个）与它匹配，如果有一个'['那么就要有一个']'与它匹配.。如果都满足，输出”YES“，else 输出”NO“. 思路； 建立一个栈，左括号’[‘,'('都入栈，右括号')',']'都出栈依次进行下去，如果栈为空，则满足条件，输出“YES”,如果不满足条件，则输出”NO“。 代码：

#include
    
     
#include
     
       using namespace std; int main() { int n; scanf(%d,&n); getchar(); while(n--) { char a[130]; stack
      
        s; gets(a); int flag=0; for(int i=0;a[i]!='';i++) { if(a[i]=='('||a[i]=='[') s.push(a[i]); else if(a[i]==')') { if(!s.empty()&&s.top()=='(') s.pop(); else {flag=1;break;} } else if(a[i]==']') { if(!s.empty()&&s.top()=='[') s.pop(); else {flag=1;break;} } } if(flag||!s.empty()) printf(No ); else printf(Yes ); } return 0; } 
      
     
    

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