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 <iostream>
#include<bits/stdc++.h>
#define F first
#define S second
using namespace std;
typedef pair<int,int> P;
vector<P> v,v2;
string s;
int n,a,b,ok,ok2,A,B,cnt1,cnt2;
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>s;
        a=0;
        b=0;
        for(int j=0;j<s.size();j++){
            if(s[j]=='('){
                cnt1++;
                a++;
            }
            else{
                cnt2++;
                if(a>0)a--;
                else b++;
            }
        }
        if(a>=b)v.push_back({b,a});
        else v2.push_back({a,b});
    }
    sort(v.begin(),v.end());
    sort(v2.begin(),v2.end());
    reverse(v2.begin(),v2.end());
    for(auto x:v){
        A-=x.F;
        if(A<0){cout<<"No"<<endl;return 0;}
        A+=x.S;
    }
    for(auto x:v2){
        A-=x.S;
        if(A<0){cout<<"No"<<endl;return 0;}
        A+=x.F;
    }
    if(A==0)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
}
