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 rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
string s[10];
int main(){
    ll n;
    cin>>n;
    rep(i,n) cin>>s[i];
    vector<P> pl,mi;
    int z=1,SUM=0;
    rep(i,n)rep(j,n){
        if(s[i][j]=='(') SUM++;
        else SUM--;
    }
    if(SUM!=0) z=0;
    rep(i,n){
        ll sum=0,mok=0;
        rep(j,s[i].size()){
            
            if(s[i][j]=='(') sum++;
            else sum--;
            mok=min(mok,sum);
        }
        if(sum>=0){
            pl.push_back(P(mok,sum));
        }
        else{
            mok=mok*(-1);
            sum=sum*(-1);
            mi.push_back(P(mok-sum,sum));
        }
    }
    ll sum=0;
    sort(all(pl),greater<P>());
    sort(all(mi));
    rep(i,pl.size()){
        if(sum<pl[i].first*(-1)) z=0;
        sum+=pl[i].second;
    }
    rep(i,mi.size()){
        if(sum<mi[i].first+mi[i].second) z=0;
        sum-=mi[i].second;
    }
    if(z) cout<<"Yes";
    else cout<<"No";
}