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;
 
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;i<n;i++)
#define ll long long
 
#define pi pair < ll , ll >
#define pb push_back
#define mp make_pair
#define F first
#define S second
 
#define fastIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
 
#define debug(x) cout<<#x<<" :: "<<x<<"\n";
#define debug2(x,y) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\n";
#define debug3(x,y,z) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<"\n";
using namespace std;

void solve() {
    ll n, m, i, j, k;
    cin>>n;
    string s;
    vector < pair < ll , string > > pr;
    fo(i,n) {
        cin>>s;
        ll cnt1 = 0, cnt2 = 0, val = 0;
        for(auto x:s) {
            if(x=='(')val++;
            else val--;
            if(val>0)cnt1 = max(cnt1, val);
            else cnt2 = max(cnt2, -1ll * val);
        }
        if(cnt1>cnt2)pr.pb({cnt1, s});
        else pr.pb({-1ll * cnt2, s});
    } 
    sort(pr.begin(), pr.end());
    reverse(pr.begin(), pr.end());
    s.clear();
    fo(i,n) for(auto x:pr[i].S)s.pb(x);
    ll val = 0;string ans = "Yes";
    for(auto x:s) {
        if(x=='(')val++;
        else val--;
        if(val<0){ans="No";break;}
    }
    if(val!=0)ans="No";
    cout<<ans;
}   

int main()
{
    fastIO;ll t=1;
    // cin>>t;
    while(t--)solve();
}