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 REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define ALLOF(c) (c).begin(), (c).end()
typedef long long ll;
typedef unsigned long long ull;

struct ST {
  string s;
  int last;
  int mn;
};
bool operator<(const ST& a, const ST& b){
  return a.mn > b.mn;
}

int main(){
  int N;
  cin >> N;
  vector<ST> v;

  int sum = 0;
  rep(i,N){
    ST st;
    cin >> st.s;
    int mn = 0;
    int now = 0;
    rep(j,st.s.size()){
      if(st.s[j] == ')') now--;
      else now++;
      mn = min(mn, now);
    }
    st.last = now;
    st.mn = mn;
    v.push_back(st);
    sum += st.last;
  }

  if(sum != 0){
    cout << "No" << endl;
    return 0;
  }
  
  sort(ALLOF(v));

  int mn = 0;
  int now = 0;
  rep(i,N){
    rep(j,v[i].s.size()){
      if(v[i].s[j] == ')') now--;
      else now++;
      mn = min(mn, now);
    }
  }
  
  if(mn >= 0 && now == 0){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
  }
  return 0;
}