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 <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <fstream>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

bool cmp(pii x, pii y) {
    return 1ll * x.first * y.second < 1ll * x.second * y.first;
}

int main() {
    ios::sync_with_stdio(false);
    int n, m = 0;
    pii a[1000006] = {};
    string s;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> s;
        int x = 0, y = 0;
        for (int j = 0; j < s.length(); j++) {
            if (s[j] == '(') x++;
            else x--;
            y = min(y, x);
        }
        a[i] = {-y, x};
    }
    sort(a, a + n, cmp);
    //for (int i = 0; i < n; i++) cout << a[i].first << ' ' << a[i].second << '\n';
    for (int i = 0; i < n; i++) {
        if (m < a[i].first) {
            cout << "No";
            return 0;
        }
        m += a[i].second;
    }
    if (m) cout << "No";
    else cout << "Yes";
}
