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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

struct P {
    bool operator<(const P &p) const {
        return c > p.c;
    }
    int d, m, c;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    vector<P> p(n);
    string s;

    constexpr int INF = 1 << 30;
    for (int i = 0; i < n; i++) {
        int d = 0, m = 0;
        cin >> s;
        for (const auto &c : s) {
            d += c == '(' ? 1 : -1;
            m = min(m, d);
        }
        p[i] = { d, m, d > 0 ? INF + m : -INF + (d - m) };
    }
    sort(p.begin(), p.end());

    int b = 1;
    int sum = 0;
    for (int i = 0; i < n; i++) {
        if (sum + p[i].m < 0) b = 0;
        sum += p[i].d;
    }
    cout << (b && sum == 0 ? "Yes" : "No") << endl;

    return 0;
}