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>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll,ll>;
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep2(i,m,n) for(int (i)=(m);(i)<(n);(i)++)
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
const ll INF = 1LL<<60;
const ll MOD = 1e9+7;


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;cin>>n;
    vector<string> s(n);rep(i,n)cin>>s[i];
    //上がるやつを下がりが小さい順に使う
    //下がるやつを下がりが大きい順に使う
    vector<P> duplus, duminus;
    rep(i,n){
        string str = s[i];
        ll u = 0, d = 0;
        rep(j,str.size()){
            if(str[j]=='(')u+=1;
            else{
                u-=1;
                d=min(d,u);
            }
        }
        if(u>=0)duplus.push_back({d,u});
        else duminus.push_back({d,u});
    }
    sort(ALL(duplus));
    sort(rALL(duminus));
    ll pos = 0;

    for(P p:duplus){
        //cout<<p.first<<","<<p.second<<endl;
        if(pos+p.first<0){
            cout<<"No";
            return 0;
        }
        pos+=p.second;
        //cout<<pos<<endl;
    }
    for(P p:duminus){
        //cout<<p.first<<","<<p.second<<endl;
        if(pos+p.first<0){
            cout<<"No";
            return 0;
        }
        pos+=p.second;
        //cout<<pos<<endl;
    }
    if(pos==0)cout<<"Yes";
    else cout<<"No";

    return 0;
}