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;
typedef long long ll;

vector<pair<int,int> > p;
bool cmp(pair<int,int> p1,pair<int,int> p2){
    return min(p1.second,p1.first+p2.second)>min(p2.second,p2.first+p1.second);
}
int main(){
    int n,sum=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        string s;
        int cnt=0,mn=0;
        cin>>s;
        for(int j=0;j<s.length();j++){
            char c=s[j];
            if(c=='(') cnt++;
            else cnt--;
            mn=min(mn,cnt);
        }
        p.push_back({cnt,mn});
        sum+=cnt;
    }
    if(sum!=0) {
        printf("No");
        return 0;
    }
    sort(p.begin(),p.end(),cmp);
    int res=0;
    for(int i=0;i<p.size();i++){
        if(res< -p[i].second){
            printf("No");
            return 0;
        }
        res+=p[i].first;
    }
    printf("Yes");
}
