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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n; cin>>n;
	vector<string> s(n);
	vector<pair<int,int>> a(n);
	rep(i,n){
		cin>>s[i];
		int cnt=0,mn=0;
		for(char c:s[i]){
			if(c=='(') cnt++;
			else       cnt--, mn=min(mn,cnt);
		}
		a[i]={cnt,min(0,cnt)-mn};
	}

	vector<int> ord(n);
	iota(ord.begin(),ord.end(),0);

	sort(ord.begin(),ord.end(),[&](int i,int j){
		int ti=(a[i].first>=0?0:1);
		int tj=(a[j].first>=0?0:1);
		if(ti!=tj) return ti<tj;
		if(ti==0) return a[i].second<a[j].second;
		else      return a[i].second>a[j].second;
	});

	int cnt=0;
	for(int i:ord){
		for(char c:s[i]){
			if(c=='(') cnt++;
			else       cnt--;
			if(cnt<0) return puts("No"),0;
		}
	}
	puts(cnt==0?"Yes":"No");

	return 0;
}
