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: Runtime Error)

#include <iostream>
#include <bits/stdc++.h>
#include <vector>
#include <cstring>
#include <climits>
using namespace std;

#define rep(i,a,b) for(int i = a;i<b;i++)

bool compare(pair<int,int> &a,pair<int,int> &b)
{
	return a.second >= b.first;
}

int main()
{
	int n;
	cin >> n;
	vector<string> s;
	s.resize(n);
	pair<int,int> *arr = new pair<int,int>[n];
	rep(j,0,n)
	{
		cin >> s[j];
		string k = s[j];
		int cl = 0, op = 0;
		rep(i,0,k.size())
		{
			if(k[i] == '(') op++;
			else if(k[i] == ')')
			{
				op--;
				if(op < 0)
				{
					cl += -op;
					op = 0;
				}
			}
		}
		arr[j].first = cl;
		arr[j].second = op;
	}

	sort(arr, arr+n, compare);
	if(arr[0].first != 0 && arr[n-1].second != 0)
	{
		cout << "No" << endl;
		return 0;
	}
	int finalAns = 0;
	rep(i,0,n-1)
	{
		finalAns += arr[i].second - arr[i+1].first;
		if(finalAns < 0) break;
	}
	if(finalAns == 0) cout << "Yes" << endl;
	else cout << "No" << endl;
	delete []arr; 
	return 0;
}