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 <cstdio>
#include <algorithm>
#include <vector>
#define PR pair<int,int>
#define F first
#define S second
using namespace std;

int n, l, r, cl, cr;
string s;
vector<PR> vt;

bool cmp(PR p, PR q) {return p.F - p.S > q.F - q.S;}

int main() {
	int i;
	cin >> n;
	while(n--) {
		cin >> s;
		l = r = 0;
		for(i=0; s[i]; i++) {
			if(s[i] == '(') l++;
			else {
				if(l) l--;
				else r++;
			}
		}
		if(l && !r) cl += l;
		else if(!l && r) cr += r;
		else if(l && r) vt.push_back({l, r});
	}
	sort(vt.begin(), vt.end(), cmp);
	for(i=0; i<vt.size(); i++) {
		if(cl < vt[i].S) return puts("No"),0;
		cl -= vt[i].S - vt[i].F;
	}
	puts(cl==cr ? "Yes" : "No");
	return 0;
}