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 ll = long long;
using P = pair<int, int>;

bool check(vector<P> A) {
    int a = 0;
    rep(i, A.size()) {
        int b = A[i].first;
        if (a+b < 0) return false;
        a += A[i].second;
    }
    return true;
}

int main() {
    int n;
    cin >> n;
    vector<P> ls, rs;
    int t = 0;
    rep(i, n) {
        string s;
        cin >> s;
        int a = 0, b = 0;
        rep(j, s.size()) {
            if (s[j] == '(') ++a;
            else --a;
            b = min(a, b);
        }
        if (a > 0) ls.emplace_back(b, a);
        else rs.emplace_back(b-a, -a);
        t += a;
    }
    sort(ls.rbegin(), ls.rend());
    sort(rs.rbegin(), rs.rend());
    if (check(ls) && check(rs) && t == 0) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}