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;

using ll = long long;
using pll = pair<ll, ll>;

int main()
{
    int N; cin >> N;

    vector<pll> b(N); // first:min, second:balance 

    for (int i = 0; i < N; i++) {
        string S; cin >> S;
        for (const auto& ch : S) {
            if (ch == '(') {
                b[i].second++;
            } else {
                b[i].second--;
                b[i].first = min(b[i].first, b[i].second);
            }
        }
    }

    sort(b.begin(), b.end(), [](const pll& a, const pll& b) {
        bool sa = signbit(a.second), sb = signbit(b.second);
        if (sa == sb) {
            if (sa == 0) {
                return a.first > b.first;
            } else {
                return a.second - a.first > b.second - b.first;
            }
        }
        return sa < sb;
    });

    int s = 0;
    for (int i = 0; i < N; i++) {
        if (s + b[i].first < 0) {
            s = -1;
            break;
        }
        s += b[i].second;
    }
    cout << (s==0 ? "Yes" : "No") << endl;
}