Problem: There is a tree with N vertices numbered 1 through N. The i-th of the N-1 edges connects vertices a_i and b_i.

Initially, each edge is painted blue. Takahashi will convert this blue tree into a red tree, by performing the following operation N-1 times:

* Select a simple path that consists of only blue edges, and remove one of those edges.
* Then, span a new red edge between the two endpoints of the selected path.



His objective is to obtain a tree that has a red edge connecting vertices c_i and d_i, for each i.

Determine whether this is achievable.

Constraints

* 2 ≤ N ≤ 10^5
* 1 ≤ a_i,b_i,c_i,d_i ≤ N
* a_i ≠ b_i
* c_i ≠ d_i
* Both input graphs are trees.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_{N-1} d_{N-1}


Output

Print `YES` if the objective is achievable; print `NO` otherwise.

Examples

Input

3
1 2
2 3
1 3
3 2


Output

YES


Input

5
1 2
2 3
3 4
4 5
3 4
2 4
1 4
1 5


Output

YES


Input

6
1 2
3 5
4 6
1 6
5 1
5 3
1 4
2 6
4 3
5 6


Output

NO

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
const int N=100005;
multiset<int> S[N];
int n,x,y;
map<pair<int,int>,int>Map;
queue<pair<int,int> > Q;
void add(int x,int y){
	S[x].insert(y);S[y].insert(x);
	if (x>y)swap(x,y);
	Map[mp(x,y)]++;
	if (Map[mp(x,y)]==2)Q.push(mp(x,y));
}
void del(int x,int y){
	S[x].erase(S[x].find(y));
	S[y].erase(S[y].find(x));
	if (x>y)swap(x,y);
	Map[mp(x,y)]--;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=2*n-2;i++){
		scanf("%d%d",&x,&y);
		add(x,y);
	}
	for (int i=1;i<n;i++){
		while (!Q.empty()&&Map[Q.front()]<2)Q.pop();
		if (Q.empty()){
			puts("NO");
			return 0;
		}
		int x=(Q.front()).fi,y=(Q.front()).se;Q.pop();
		del(x,y);
		del(x,y);
		if (S[x].size()<S[y].size())swap(x,y);
		vector<int> v;
		for (int j:S[y])v.push_back(j);
		for (int j:v){
			del(y,j);
			add(x,j);
		}
	}
	puts("YES");
}