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;
typedef double db;
string s[1000005];
vector<pair<int,int>> a,b;
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>s[i];
    }
    for(int i=1;i<=n;i++){
        int l=0,r=0;
        for(int j=0;j<s[i].length();j++){
            if(s[i][j]=='('){
                l++;
            }
            else{
                if(l>0) l--;
                else r++;
            }
        }
        if(l<=r) a.push_back({l,r});
        else b.push_back({r,l});
    }
    sort(a.begin(),a.end());
    sort(b.begin(),b.end());
    reverse(b.begin(),b.end());
    int sum=0;
    for(int i=0;i<a.size();i++){
        sum-=a[i].first;
        if(sum<0){
            cout<<"No"<<endl;return 0;
        }
        sum+=a[i].second;
    }
    for(int i=0;i<b.size();i++){
        sum-=b[i].second;
        if(sum<0){
            cout<<"No"<<endl;return 0;
        }
        sum+=b[i].first;
    }
    if(sum==0) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;

}