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;
#define rep(i,n) for(int i=0; i<(n); i++)
constexpr int MX = 1000005;
int n, el, er;
char s[MX];
using P = pair<int,int>;
P ls[MX], rs[MX];
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%s", s);
		int len = strlen(s), l = 0, r = 0;
		rep(j,len){
			if(s[j] == '(') r++;
			else if(r > 0) r--;
			else l++;
		}
		if(l < r) ls[el++] = {l,r};
		else rs[er++] = {l,r};
	}
	sort(ls, ls+el);
	sort(rs, rs+er, [](P &l, P &r){return l.second > r.second;});
	auto f = [](P l, P r){return P{l.first + max(0,r.first-l.second), r.second + max(0, l.second - r.first)};};
	P p = accumulate(ls, ls+el, P{0,0}, f);
	p = accumulate(rs, rs+er, p, f);
	puts(p == P{0,0} ? "Yes" : "No");
}