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 <bits/stdc++.h>
#define pb push_back
#define mk make_pair
#define l first
#define r second
using namespace std;
typedef long long ll;
typedef pair <int,int> PAIR;
const int N = 1e6+5;
const ll M = 998244353;

struct node{
	ll l,r;
	node(){}
	node(ll l,ll r):l(l),r(r){}
} a[N];
inline node cal(string s){
	ll L = 0,R = 0;
	for(int i = 0;i < s.size();++i){
		if(s[i] == '(') L++;
		else{
			if(L) L--;
			else R++;
		}
	}
	return node(L,R);
}

bool comp(node x,node y){
	if(x.r == y.r) return x.l > y.l;
	return x.r < y.r;
}

int main() {
	int n;
	string s;
	cin >> n;
	for(int i = 1;i <= n;++i){
		cin >> s;
		a[i] = cal(s);
	}
	sort(a+1,a+n+1,comp);
	
	ll cnt = 0;
	bool flag = 1;
	for(int i = 1;i <= n;++i){
		if(a[i].r) cnt -= a[i].r;
		if(cnt < 0) flag = 0;
		
		cnt += a[i].l;
	}
	
	if(!cnt && flag) printf("Yes\n");
	else printf("No\n");
	return 0;
}