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<bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define int ll
#define pb push_back
#define setbits(x)          __builtin_popcountll(x)
#define endl "\n"
typedef long long ll;
int32_t main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin>>n;
pair<int,string > arr[n+1];
for(int i=0;i<n;i++)
{
    cin>>arr[i].ss;
    string str=arr[i].ss;
    int x=str.size();
    int mneg=0;
    int c=0;
    for(int j=0;j<x;j++)
    {
        if(str[i]=='(')
            c+=1;
        else
            c-=1;
        mneg=min(mneg,c);
    }
    arr[i].ff=mneg+c;
}
sort(arr,arr+n);
string res="";
for(int i=n-1;i>=0;i--)
    res+=arr[i].ss;
int c=0;
for(int i=0;i<res.size();i++)
{
    if(res[i]=='(')
    {
        c+=1;
    }
    else
        c-=1;
    
}
if(c!=0)
{
    cout<<"No"<<endl;
    return 0;
}
cout<<"Yes"<<endl;
    return 0;
}