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>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <queue>

using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<n;++i)
#define mod (ll)(1e9+7)
#define FIX(a) ((a)%mod+mod)%mod
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define fi first
#define se second
#define pb push_back
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<pair<ll,pair<ll,ll> > > rl(n);
    REP(i,n){
        string s;
        cin >> s;
        ll l=0,r=0;
        for(char c:s){
            if(c=='('){
                r++;
            }else{
                if(r>0){
                    r--;
                }else{
                    l++;
                }
            }
        }
        if(l==0){
            rl[i] = make_pair(-2,make_pair(l,r));
        }else if(r==0){
            rl[i] = make_pair(2,make_pair(l,r));            
        }else if(r>=l){
            rl[i] = make_pair(-1,make_pair(l,r));
        }else{
            rl[i] = make_pair(1,make_pair(l,r));            
        }
    }
    sort(ALL(rl));
    ll now = 0;
    if(rl[0].se.fi>0||rl[n-1].se.se>0){
        cout << "No" << endl;
        return 0;
    }
    REP(i,n){
        ll l = rl[i].se.fi;
        ll r = rl[i].se.se;
        now -= l;
        if(now<0){
            cout << "No" << endl;
            return 0;
        }
        now += r;
    }
    if(now==0){
        cout << "Yes" << endl;
    }else{ 
        cout << "No" << endl;
    }
    return 0;
}