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 rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
const int MX = 1000005;

int main(){
    int n;
    cin >> n;
    vector<string> s(n);
    rep(i,n) cin >> s[i];
    vector<P> LR(n);
    rep(i,n){
        int l = 0;
        int r = 0;
        rep(j,s[i].size()){
            if(s[i][j] == '('){
                r ++;
            }
            else{
                if(r > 0) --r;
                else l++;
            } 
        }
        LR[i].first = l;
        LR[i].second = r;
    }

    vector<vector<int>> L(MX);
    vector<int> R(MX,0);
    rep(i,n){
        L[LR[i].first].push_back(i);
        R[LR[i].second] ++;
    }

    if(L[0].size() == 0 || R[0] == 0){
        cout << "No" << endl;
        return 0;
    }

    bool ok = true;
    rep(i,MX){
        if(L[i].size() == R[i]) ;
        else{
            ok= false;
            break;
        }
    }
    
    if(!ok){
        cout << "No" << endl;
        return 0;
    }

    vector<bool> used(n,false);
    used[L[0].back()] = true;
    int now = LR[L[0].back()].second;
    L[0].pop_back();

    while(1){
        if(L[now].size() == 0) break; 

        if(LR[L[now].back()].second == 0 && L[now].size() > 1){
            swap(L[now].back(),L[now].front());
        }

        used[L[now].back()] = true;
        int next = LR[L[now].back()].second;
        L[now].pop_back();
        now = next;
    }

    rep(i,n){
        if(!used[i]){
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes"  << endl;
    return 0;
}