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;
#define INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(0), cout << fixed << setprecision(20);
#define int long long
const int INF = 1e18, MOD = 1e9 + 7;
signed main() {
    int n;
    string s;
    vector<pair<int, int>> ss;
    cin>>n;
    for (int i = 0; i < n; i++) {
        cin>>s;
        int cnt = 0;
        int mn = INF;
        for (char c: s) {
            if (c == '(') cnt++;
            else cnt--;
            mn = min(mn, cnt);
        }
        ss.push_back({cnt, mn});
    }
    sort(ss.begin(), ss.end(), [] (pair<int, int> x, pair<int, int> y) {
        return min(x.second, x.first + y.second) > min(y.second, y.first + x.second);
    });
    int cnt = 0;
    for (pair<int, int> p: ss) {
        if (cnt + p.second < 0) {
            cout<<"No"<<endl;
            return 0;
        }
        cnt += p.first;
    }
    if (cnt == 0) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
}