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
/*
 *  Language:   C++
 *  Author:     Debajyoti Dasgupta
 */
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

bool check(const vector<pair<int, int>>& a) {
    int cur = 0;
    for (auto i:a) {
        if (cur + i.first < 0) return false;
        cur += i.second;
    }
    return true;
}

int main() {
    int n, total = 0;
    cin >> n;
    vector<pair<int, int>> ls, rs;
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        int mn = 0, cur = 0;
        for (char &s1:s) {
            cur += (s1 == '(' ? 1 : -1);
            mn = min(mn, cur);
        }
        total += cur;
        if (cur > 0) {
            ls.emplace_back(mn, cur);
        } else {
            rs.emplace_back(mn - cur, -cur);
        }
    }
    sort(ls.rbegin(), ls.rend());
    sort(rs.rbegin(), rs.rend());
    if (!total && check(ls) && check(rs)) {
        cout << "Yes";
    } else {
        cout << "No";
    }
    return 0;
}
