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;
#define rep(i,n)for(int i=0;i<(n);i++)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
using ll = long long;
using P = pair<int, int>;

bool check(vector<P> vec){
    int h = 0;
    for(auto p : vec){
        int b = h + p.first;
        if(b < 0) return false;
        h += p.second;
    }
    return true;
}

int main(){
    int n; cin >> n;
    vector<P> ls, rs;
    int total = 0;
    rep(i,n){
        string s; cin >> s;
        int h = 0, b = 0;//h:増減,b:最下点
        for(auto c : s){
            if(c == '(') h++;
            else h--;
            b = min(b,h);
        }
        if(h > 0)ls.emplace_back(b, h);
        else rs.emplace_back(b-h, -h);
        total += h;
    }
    sort(RALL(ls));
    sort(RALL(rs));
    if(check(ls) && check(rs) && total == 0) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}