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;
typedef pair<int,int> ii;
ii a[1000005];
bool cmp(ii a, ii b){
    int dif1 = a.second-a.first;
    int dif2 = b.second-b.first;
    if ( (dif1 < 0) != (dif2 < 0) ) return (dif1 > 0);
    if (dif1 < 0 && dif2 < 0){
        if (a.first != b.first) return a.first > b.first;
        else return a.second-a.first > b.second-b.first;
    }
    else {
        if (a.first != b.first) return a.first < b.first;
        else return a.second-a.first > b.second-b.first;
    }
}
int main(){
    int n;
    scanf("%d",&n);
    for (int i = 0; i < n; i++){
        string s;
        cin >> s;
        int cur = 0;
        for (int j = 0; j < s.size(); j++){
            if (s[j] == '(') cur++;
            else{
                if (cur == 0) a[i].first++;
                else cur--;
            }
            a[i].second = cur;
        }
    }
    sort(a,a+n,cmp);
    int X = 0, Y = 0;
    for (int i = 0; i < n; i++){
        //printf("%d %d\n",a[i].first,a[i].second);
        if (a[i].first > X){
            printf("No");
            return 0;
        }
        X -= a[i].first;
        X += a[i].second;
    }
    printf("%s\n",X == 0 ? "Yes":"No");
}