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;
int n,cc;
map<long long,int>mp;
set<int>g[100005];
queue<pair<int,int> >q;
int main()
{
	scanf("%d",&n);
	for(int i=0;i<2*n-2;i++)
	{
		int u,v;scanf("%d%d",&u,&v);
		mp[(long long)u*n+v]++;mp[(long long)v*n+u]++;
		g[u].insert(v);g[v].insert(u);
		if(mp[(long long)u*n+v]==2)q.push(make_pair(u,v));
	}
	while(!q.empty())
	{
		int u=q.front().first,v=q.front().second;q.pop();
		if(mp[(long long)u*n+v]!=2)continue;
		cc++;
		if(g[u].size()<g[v].size())swap(u,v);
		while(g[v].size())
		{
			int w=*g[v].begin();
			mp[(long long)u*n+w]+=mp[(long long)v*n+w];
			mp[(long long)w*n+u]+=mp[(long long)w*n+v];
			mp[(long long)v*n+w]=mp[(long long)w*n+v]=0;
			g[u].insert(w);g[w].insert(u);g[v].erase(w);g[w].erase(v);
			if(mp[(long long)u*n+w]==2)q.push(make_pair(u,w));
		}
	}
	if(cc==n-1)puts("YES");else puts("NO");
	return 0;
}