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;

string remv(const string &S) {
    string res = "";
    for(auto &c : S) {
        if(res.empty()) res += c;
        else if(res.back() == '(' and c == ')') res.pop_back();
        else res += c;
    }
    return res;
}

int main() {
    int N; cin >> N;
    using Bracket = tuple<string, int, int>;
    vector<Bracket> L, R;
    for(int i = 0; i < N; ++i) {
        string S; cin >> S;
        S = remv(S);
        int l = count(S.begin(), S.end(), '(');
        int r = count(S.begin(), S.end(), ')');
        if(l >= r) L.emplace_back(S, l, r);
        else R.emplace_back(S, l, r);
    }
    sort(L.begin(), L.end(), [](const Bracket &A, const Bracket &B){
        return get<2>(A) < get<2>(B);
    });
    sort(R.begin(), R.end(), [](const Bracket &A, const Bracket &B){
        return get<1>(A) > get<1>(B);
    });
    string ans = "";
    for(auto &[s, l, r] : L) ans += s;
    for(auto &[s, l, r] : R) ans += s;
    cout << (remv(ans).empty() ? "Yes" : "No") << '\n';
    return 0;
}
