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;

int a[1111111], b[1111111];
int main() {
    int n, S = 0, T = 0, D = 0;
    cin >> n;
    for(int i = 0; i < n; i++) {
        string z;
        cin >> z;
        int d = 0, s = 0;
        for(auto c : z)
            if(c == '(')
                d++;
            else
                d--, s = max(s, -d);
        if(d > 0)
            a[s] += d, S = max(S, s);
        else
            s = max(s + d, 0), b[s] += d, T = max(T, s);
        D += d;
    }
    int x = 0;
    if(D)
        goto no;
    for(int s = 0; s <= S; s++) {
        if(s > x)
            goto no;
        x += a[s];
    }
    x = 0;
    for(int s = 0; s <= T; s++) {
        if(s > x)
            goto no;
        x -= b[s];
    }
    cout << "Yes" << endl;
    return 0;
no:
    cout << "No" << endl;
    return 0;
}
