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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

bool check(vector<pair<ll, ll>> s){
    ll h = 0;
    for (auto p : s){
        ll b = h + p.first;
        if (b < 0) return false;
        h += p.second;
    }
    return true;
}

int main() {
    ll N, sum = 0;
    cin >> N;
    vector<pair<ll, ll>> up, down;
    rep(i, N) {
        string s;
        cin >> s;
        int n = s.size();
        int mini = 0;
        int diff = 0;
        for (int j = 0; j < n; j++) {
            if (s[j] == '(') {
                diff++;
            } else {
                diff--;
            }
            mini = min(mini, diff);
        }
        if (diff > 0) {
            up.emplace_back(mini, diff);
        } else {
            down.emplace_back(mini-diff, -diff);
        }
        sum += diff;
    }
    sort(up.rbegin(), up.rend());
    sort(down.rbegin(), down.rend());
    puts(check(up) && check(down) && sum == 0 ? "Yes" : "No");
}