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;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    vector<array<int, 2>> v, w;
    for (int i = 0; i < n; i++) {
        string s; cin >> s;
        int mini = 0; int curr = 0;
        for (auto &c: s) {
            if (c == '(') curr++;
            else curr--;
            mini = min(mini, curr);
        }
        if (curr >= 0) v.push_back({mini, curr});
        else w.push_back({mini-curr, -curr});
    }
    sort(v.rbegin(), v.rend());
    sort(w.rbegin(), w.rend());
    int curr = 0;
    bool ok = true;
    for (auto &[m, c]: v) {
        if (curr + m < 0) ok = false;
        curr += c;
    }
    int prev = 0;
    for (auto &[m, c]: w) {
        if (prev + m < 0) ok = false;
        prev += c;
    }
    if (curr != prev) ok = false;
    if (ok) cout << "Yes\n";
    else cout << "No\n";
    return 0;
}