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 <set> 
#include <cstdio>
#include <iostream>
#define N 100010
using namespace std;
set<int> v[N];
pair<int,int> Q[N];
int n,head,tail,fa[N];
int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }
int main()
{
	scanf("%d",&n);
	for (int i=1,x,y;i<=n*2-2;i++) 
	{
		scanf("%d%d",&x,&y);
		if (!v[x].count(y)) v[x].insert(y),v[y].insert(x); else Q[tail++]=make_pair(x,y);
	} 
	for (int i=1;i<=n;i++) fa[i]=i;
	while (head<tail)
	{
		int x=find(Q[head].first),y=find(Q[head++].second);
		if (v[x].size()<v[y].size()) swap(x,y);
		for (set<int>::iterator it=v[y].begin();it!=v[y].end();it++) if (*it!=x)
		{
			int l=x,r=*it;
			v[r].erase(y);
			if (!v[l].count(r)) v[l].insert(r),v[r].insert(l); else Q[tail++]=make_pair(l,r);
		}
		v[x].erase(y),fa[y]=x;
	}
	if (tail==n-1) printf("YES"); else printf("NO");
}