Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <inttypes.h>

int main(void){
	int n;
	char s[1000001]={'\0'};
	int i;
	int len;
	int r=0;
	int l=0;
	int j;
	int rflag1=0,rflag2=0;
	int lflag1=0,lflag2=0;
	int lans=0,rans=0;
	int kakko=0;

	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%s",s);
		len=strlen(s);
		for(j=0,l=0,r=0,lflag1=0,rflag1=0;j<len;j++){
			if(s[j]=='('){
				l++;
				lans++;
			}else{
				r++;
				rans++;
			}
			if(j==0){
				if(s[j]=='('){
					lflag1=1;
				}
			}
		}
		if(lflag1==1){
			if(l==r){
				if(s[len-1]==')'){
					lflag1=0;
					kakko++;
				}
			}
		}else{
			if(s[len-1]==')'){
				rflag1=1;
			}
		}
		if(lflag1==1){
			lflag2++;;
		}
		if(rflag1==1){
			rflag2++;;
		}
	}
	if(lans==rans){
		if(lflag2==rflag2&&(lflag2>0||kakko==n)){
			printf("Yes");
			return 0;
		}
	}
	printf("No");

	return 0;
}