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 <cstdint>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <string>
#include <utility>
#include <vector>

int main() {
  size_t n;
  scanf("%zu", &n);

  std::vector<std::pair<int, int>> p(n);
  char buf[1048576];
  for (auto& pi: p) {
    scanf("%s", buf);
    std::string s = buf;
    int diff = 0;
    int depth = 0;
    for (auto c: s) {
      diff += ((c == '(')? +1: -1);
      depth = std::min(depth, diff);
    }
    pi = std::make_pair(depth, diff);
  }

  std::sort(p.begin(), p.end(), [](auto const& p1, auto const& p2) {
    auto [de1, di1] = p1;
    auto [de2, di2] = p2;
    if (de1 != de2) return de1 > de2;
    return di1 < di2;
  });

  int diff = 0;
  for (auto const& [de, di]: p) {
    if (diff+de < 0) return puts("No"), 0;
    diff += di;
  }

  if (diff != 0) return puts("No"), 0;
  puts("Yes");
}