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;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/f.in", "r", stdin);
//  freopen("../IO/f.out", "w", stdout);
#else
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
#endif
  solve(std::cin, std::cout);
  return 0;
}

constexpr int N = 1000'0225;
char buff[N];

void solve(std::istream &in, std::ostream &out) {
  int n;
  in >> n;
  vector<pair<int, int>> sq;
  for (int i = 0; i < n; ++i) {
    in >> buff;
    int len = strlen(buff);
    int mn = len, bal = 0;
    for (int j = 0; j < len; ++j) {
      if (buff[j] == '(') {
        ++bal;
      } else {
        --bal;
      }
      mn = min(mn, bal);
    }
    if (mn == 0 && bal == 0) continue;
    if (mn < 0) {
      sq.emplace_back(-mn, bal - mn);
    } else {
      sq.emplace_back(0, bal);
    }
  }
  if (sq.empty()) {
    out << "Yes\n";
    return;
  }
  int64_t suml = 0, sumr = 0;
  for (auto pr : sq) {
    suml += pr.first, sumr += pr.second;
  }
  if (suml != sumr) {
    cout << "No\n";
    return;
  }
  sort(sq.begin(), sq.end());
  sumr = 0, suml = 0;
  bool bad = false;
  for (auto pr : sq) {
    sumr -= pr.first;
    bad |= sumr < 0;
    sumr += pr.second;
  }
  if (bad || sumr != 0) {
    out << "No\n";
  } else {
    out << "Yes\n";
  }
}