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 <cassert>
#include <iostream>
using namespace std;

bool cmp (const pair<int, int> &lhs, const pair<int, int> &rhs) {
    if (lhs.second >= 0) {
        if (rhs.second >= 0) return lhs.first > rhs.first;
        else return true;
    } else {
        if (rhs.second >= 0) return false;
        else return lhs.second - lhs.first > rhs.second - rhs.first;
    }
}

int main() {
    int n; cin >> n;
    pair<int, int> a[n];
    int sum = 0;
    int sum_len = 0;
    for (int i = 0; i < n; i++) {
        string str; cin >> str;
        int s = 0;
        a[i] = {0, 0};
        for (char c : str) {
            if (c == '(') s++;
            else s--;
            a[i].first = min(a[i].first, s);
        }
        a[i].second = s;
        sum += s;
        sum_len += str.size();
    }
    if (sum != 0) { cout << "No\n"; return 0; }
    sort(a, a+n, cmp);
    sum = 0;
    for (auto &p : a) {
        if (sum + p.first < 0) { cout << "No\n"; return 0; }
        else sum += p.second;
    }
    cout << "Yes\n";
}
