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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n) - 1; i >= 0; i--)
#define range(a) a.begin(), a.end()

pair<int, int> f(string s) {
    int x = 0;
    int res = 0;
    for (char c : s) {
        if (c == '(') x++;
        if (c == ')') x--;
        res = min(res, x);
    }
    return make_pair(res, x);
}

int main() {
    int N; cin >> N;
    vector<string> S(N); rep(i, N) cin >> S[i];
    vector<int> F(N), G(N);
    rep(i, N) {
        auto tmp = f(S[i]);
        F[i] = tmp.first;
        G[i] = tmp.second;
    }
    vector<int> ord(N); rep(i, N) ord[i] = i;
    sort(range(ord), [&](int i, int j) {
        int ij = min(F[i], G[i] + F[j]);
        int ji = min(F[j], G[j] + F[i]);
        if (ij != ji) return ij > ji;
        return G[i] > G[j];
    });
    string ans;
    for (int i : ord) ans += S[i];
    auto tmp = f(ans);    
    cout << (tmp.first == 0 and tmp.second == 0 ? "Yes" : "No") << endl;
}