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: Accepted)

#include <cstdio>
#include <algorithm>
#define maxn 1000010
using namespace std;
char s[maxn];
struct node{
	int l,r;
}a[maxn];
int tot;
int cmp(node a,node b){
	return a.l-a.r>b.l-b.r;
}
int main(){
	int n,i,j,l,r,ln,rn;
	ln=rn=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		l=r=0;
		scanf("%s",s+1);
		for(j=1;s[j];j++){
			if(s[j]=='(')l++;
			else if(s[j]==')'){
				if(l)l--;
				else r++;
			}
		}
		if(l&&!r)ln+=l;
		else if(!l&&r)rn+=r;
		else if(l&&r) a[++tot]=(node){l,r};
	}
	sort(a+1,a+1+tot,cmp);
	for(i=1;i<=tot;i++)
		if(ln<a[i].r)return printf("No\n"),0;
		else ln-=a[i].r,ln+=a[i].l;
	if(ln!=rn)return printf("No\n"),0;
	printf("Yes\n");
	return 0;
}