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;
using ll=long long;
constexpr ll mod=1000000007;
constexpr ll mod998=998244353;


int N;
string S[1000000];
vector<pair<int,int>>inc,de;

int main(){
	cin>>N;
	for(int i=0;i<N;i++)cin>>S[i];
	int cnt=0;
	for(int i=0;i<N;i++){
		for(char c:S[i]){
			if(c=='(')cnt++;
			else cnt--;
		}
	}
	if(cnt!=0){
		puts("No");
		return 0;
	}
	for(int i=0;i<N;i++){
		int cc=0;
		int mn=0;
		for(char c:S[i]){
			if(c=='(')cc++;
			else cc--;
			mn=min(mn,cc);
		}
		if(cc>=0){
			inc.push_back({-mn,cc});
		}
		else{
			de.push_back({cc-mn,-cc});
		}
	}
	sort(inc.begin(),inc.end());
	sort(de.begin(),de.end());
	int t=0;
	for(auto p:inc){
		if(p.first>t){
			puts("No");
			return 0;
		}
		t+=p.second;
	}
	t=0;
	for(auto p:de){
		if(p.first>t){
			puts("No");
			return 0;
		}
		t+=p.second;
	}
	puts("Yes");
}
