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;
const int N = 1e6 + 10;
typedef pair<int, int> PII;

vector<PII> le, ri;
char s[N];

int main(){
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++){
        scanf("%s", s + 1);
        int len = strlen(s + 1);
        int l = 0, r = 0;
        for(int j = 1; j <= len; j ++){
            if(s[j] == '(') l ++;
            else{
                if(l > 0) l --;
                else r ++;
            }
        }
        if(l >= r) le.push_back({r, l});
        else ri.push_back({l, r});
    }
    sort(le.begin(), le.end());
    sort(ri.begin(), ri.end(), greater<PII>());
    int cnt = 0, mark = 1;
    for(int i = 0; i < le.size(); i ++){
        cnt -= le[i].first;
        if(cnt < 0) { mark = 0; break; }
        cnt += le[i].second;
    }
    for(int i = 0; i < ri.size(); i ++){
        cnt -= ri[i].second;
        if(cnt < 0) { mark = 0; break; }
        cnt += ri[i].first;
    }
    
    if(cnt) mark = 0;
    if(mark) puts("Yes");
    else puts("No");

    return 0;
}