As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: 

Correct parentheses sequences can be defined recursively as follows: 
1.The empty string "" is a correct sequence. 
2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence. 
3.If "X" is a correct sequence, then "(X)" is a correct sequence. 
Each correct parentheses sequence can be derived using the above rules. 
Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))". 

Now Yuta has a parentheses sequence SS, and he wants Rikka to choose two different position i,ji,j and swap Si,SjSi,Sj. 

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation. 

It is too difficult for Rikka. Can you help her?

InputThe first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100 

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.OutputFor each testcase, print "Yes" or "No" in a line.Sample Input

3
4
())(
4
()()
6
)))(((

Sample Output

Yes
Yes
No

Hint

For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

水题，注意两点1一定要交换所以"()"不行；2记录 l和r 遇到第一次r>l的情况和最后面的'('交换位置，之后就不能交换位置了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<vector>
 5 #include<cstring>
 6 #include<string>
 7 #include<algorithm>
 8 #include<map>
 9 #include<cmath>
10 #include<math.h>
11 using namespace std;
12 
13 
14 
15 int main()
16 {
17     int T,n;
18     int l,r,sum;
19     scanf("%d",&T);
20     char a[100005];
21     bool flag1,flag2;
22     while(T--)
23     {
24         scanf("%d",&n);
25         if(n==0)
26         {
27             printf("Yes\n");
28             continue;
29         }
30         cin>>a;
31         l=0;r=0;sum=0;
32         if(n==2&&a[0]=='('&&a[1]==')')
33             printf("No\n");
34         else if(n%2!=0)
35             printf("No\n");
36         else
37         {
38             flag1=false;flag2=true;
39             for(int i=0;i<n;i++)
40             {
41                 if(a[i]=='(')
42                     l++;
43                 else if(a[i]==')')
44                     r++;
45                 if(r>l&&!flag1)
46                 {
47                     for(int j=n-1;j>i;j--)
48                     {
49                         if(a[j]=='(')
50                         {
51                             a[i]='(';
52                             a[j]=')';
53                             l++;
54                             r--;
55                             flag1=true;
56                             break;
57                         }
58                     }
59                 }
60                 else if(r>l&&flag1==true)
61                 {
62                     flag2=false;
63                     break;
64                 }
65             }
66             if(flag2&&l==r)
67                 printf("Yes\n");
68             else
69                 printf("No\n");
70         }
71         
72     }
73     return 0;
74 }