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 <algorithm>
#include <cstdio>
const int MAXN = 1e6+5;
char S[MAXN];
int N;

int sgn(int x) { return (x > 0) - (x < 0); }

struct Data {
  int dep, lvl;
  
  bool operator < (const Data& o) const {
    int dl = sgn(lvl);
    int dr = sgn(o.lvl);
    // +1 -> 0 -> -1
    if (dl != dr) return (dl > dr);
    // (dl == dr)
    if (dl >= 0) return (dep < o.dep);
    // (dl == -1)
    return (dep + lvl) > (o.dep + o.lvl);
  }
};
Data A[MAXN];

bool solve() {
  int lvl = 0;
  for (int i = 0; i < N; ++i) {
    if (lvl < A[i].dep) return false;
    lvl += A[i].lvl;
    if (lvl < 0) return false;
  }
  return (lvl == 0);
}

int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf(" %s", S);
    int dep = 0;
    int lvl = 0;
    for (char *ptr = S; *ptr; ++ptr) {
      if (*ptr == '(') {
        ++lvl;
      } else {
        --lvl;
      }
      dep = std::min(lvl, dep);
    }
    A[i].dep = -dep;
    A[i].lvl = lvl;
  }
  std::sort(A, A+N);
  puts(solve() ? "Yes" : "No");
  return 0;
}