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>
using namespace std;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(9), 0);
//int SEGV = getenv("D") || (exit(system("D= SEGFAULT_SIGNALS=all catchsegv ./prog.exe") >> 8), 0);
int main() {
    int N; cin >> N;
    vector<string> S(N); for (auto &s : S) cin >> s;
    vector<tuple<int, int, int>> up, notup;
    for (int i = 0; i < N; i++) {
        int lv = 0, mi = 0;
        for (int c : S[i]) {
            if (c == '(') {
                lv++;
            } else {
                lv--;
                mi = min(mi, lv);
            }
        }
        if (lv > 0) {
            up.emplace_back(-mi, mi, lv);
        } else {
            notup.emplace_back(mi - lv, mi, lv);
        }
    }
    sort(begin(up), end(up));
    sort(begin(notup), end(notup));
    int curr = 0;
    bool ans = true;
    for (auto& tup : up) {
        int dummy, mi, lv; tie(dummy, mi, lv) = tup;
        if (curr + mi < 0) ans = false;
        curr += lv;
    }
    for (auto& tup : notup) {
        int dummy, mi, lv; tie(dummy, mi, lv) = tup;
        if (curr + mi < 0) ans = false;
        curr += lv;
    }
    if (curr != 0) ans = false;
    cout << (ans ? "Yes" : "No") << '\n';
    return 0;
}