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"
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
using P = pair <int, int>;

bool check(vector<P> st){
    int bottom, last= 0;
    for(P p :st){
        bottom = last + p.first;
        if (bottom < 0) return false;
        last += p.second;
    }
    return true;
}


int main(){
    int n;
    cin >> n;
    vector<P> left, right;
    int total = 0;
    rep(i, n){
        string s;
        cin >> s;
        int bottom = 0, last = 0;
        for(char c :s){
            if (c == '(') last += 1;
            else last -= 1;
            bottom = min(bottom, last);
        }
        if (last > 0) left.emplace_back(bottom, last);
        else right.emplace_back(bottom - last, -last);
        total += last;
    }
    sort(left.rbegin(), left.rend());
    sort(right.rbegin(), right.rend());

    if (check(left) && check(right) && total == 0) cout << "Yes" << endl;
    else cout << "No" << endl;

    return 0;
} 