A bracket sequence is a string that is one of the following:

1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. 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`.

Examples

Input

2
)
(()


Output

Yes


Input

2
)(
()


Output

No


Input

4
((()))
((((((
))))))
()()()


Output

Yes


Input

3
(((
)
)


Output

No
#include <bits/stdc++.h>

using namespace std;

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
typedef long long ll;
typedef long double ld;

const int N = (int) 1e6 + 7;
int n, sumtop;

struct T {
  int top;
  int mn;
};

bool operator < (T a, T b) {
  return min(a.mn, a.top + b.mn) > min(b.mn, b.top + a.mn);
}

T a[N];

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    int top = 0, mn = 0;
    for (auto &ch : s) {
      if (ch == '(') {
        top++;
      } else {
        top--;
      }
      mn = min(mn, top);
    }
    a[i] = {top, mn};
    sumtop += top;
  }
  if (sumtop != 0) {
    cout << "No\n";
    return 0;
  }
  sort(a + 1, a + n + 1);
  int cur = 0;
  for (int i = 1; i <= n; i++) {
    if (cur + a[i].mn < 0) {
      cout << "No\n";
      return 0;
    }
    cur += a[i].top;
  }
  cout << "Yes\n";
  return 0;
}
