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<bits/stdc++.h>
using namespace std;

string s;

struct st{
    int c1,c2,mn1,mn2;
};
vector < st > L, R;

int main(){

    int j,i,n,c1,c2,mn1,mn2;
    cin>>n;
    for(i=1; i<=n; i++){
        cin>>s;
        c1=c2=mn1=mn2=0;
        for(j=0; j<s.size(); j++){
            if(s[j]=='(')c1++;
            else c1--;
            mn1=min(mn1,c1);
        }
        for(j=s.size()-1; j>=0; j--){
            if(s[j]==')')c2++;
            else c2--;
            mn2=min(mn2,c2);
        }
        if(c1>=0)L.push_back({c1,c2,mn1,mn2});
        else R.push_back({c1,c2,mn1,mn2});
    }
    sort(L.begin(),L.end(), [] (st a , st b){return a.mn1>b.mn1;});
    sort(R.begin(),R.end(), [] (st a , st b){return a.mn2<b.mn2;});

    int cnt=0;
    for(auto k:L){
        if(cnt+k.mn1<0){
            cout<<"No"<<endl;
            return 0;
        }
        cnt+=k.c1;
    }
    for(auto k:R){
        if(cnt+k.mn1<0){
            cout<<"No"<<endl;
            return 0;
        }
        cnt+=k.c1;
    }
    if(cnt)cout<<"No"<<endl;
    else cout<<"Yes"<<endl;

    return 0;
}