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 <cstdio>
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>

using namespace std;

typedef long long ll;
// const ll MOD = 1000000007;
const ll inf = 999999999999;

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

int main() {
   int n; cin >> n;
   vector<pair<int,int>> ls,rs;
   int total = 0;
   for (int i = 0; i < n; i++) {
       string s; cin >> s;
       int saikaten = 0;
       int last = 0;
       for (int i = 0; i < s.size(); i++) {
           if (s[i] == '(') last++;
           else last--;
           saikaten = min(last,saikaten);
       }
       total += last;
       if (last > 0) ls.emplace_back(saikaten, last);
       else rs.emplace_back(saikaten-last,-last);
    }
    sort(ls.rbegin(),ls.rend());
    sort(rs.rbegin(),rs.rend());
    if (check(ls) && check(rs) && total == 0) cout << "Yes\n";
    else cout << "No\n";
}