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: Accepted)

#include<bits/stdc++.h>

int main(){
    using namespace std;
    unsigned long N;
    cin >> N;
    vector<pair<long, long>> u, d;
    long check_s{0};
    for(unsigned long i{0}; i < N; ++i){
        string S;
        cin >> S;
        long t{0}, p{0};
        for(const auto& c : S)if(c & 1)p = min(p, --t);
        else++t;
        if(t >= 0)u.emplace_back(p, t);
        else d.emplace_back(p - t, -t);
        check_s += t;
    }
    if(check_s)return 0 & puts("No");
    sort(u.begin(), u.end(), greater<>());
    sort(d.begin(), d.end(), greater<>());
    {
        priority_queue<unsigned long> pq;
        long now{0}, cnt{0};
        auto it = u.begin();
        while(cnt < u.size()){
            while(it != u.end() && now + it -> first >= 0){
                pq.emplace(it -> second);
                ++it;
            }
            if(pq.empty())return 0 & puts("No");
            now += pq.top();
            pq.pop();
            ++cnt;
        }
    }
    {
        priority_queue<unsigned long> pq;
        long now{0}, cnt{0};
        auto it = d.begin();
        while(cnt < d.size()){
            while(it != d.end() && now + it -> first >= 0){
                pq.emplace(it -> second);
                ++it;
            }
            if(pq.empty())return 0 & puts("No");
            now += pq.top();
            pq.pop();
            ++cnt;
        }
    }
    puts("Yes");
    return 0;
}