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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#define mp(x,y) make_pair(x,y)
using namespace std;

typedef pair<int,int> pi;

const int N=100005;

int n,f[N];
set<int> se[N];
map<pi,int> ma;
queue<pi> que;

int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int find(int x)
{
	if (f[x]==x) return x;
	else return f[x]=find(f[x]);
}

int main()
{
	n=read();
	for (int i=1;i<=n*2-2;i++)
	{
		int x=read(),y=read();
		se[x].insert(y);se[y].insert(x);
		ma[mp(x,y)]++;ma[mp(y,x)]++;
		if (ma[mp(x,y)]==2) que.push(mp(x,y));
	}
	for (int i=1;i<=n;i++) f[i]=i;
	set<int>::iterator it;
	while (!que.empty())
	{
		int x=que.front().first,y=que.front().second;que.pop();
		x=find(x);y=find(y);
		se[x].erase(y);se[y].erase(x);
		if (se[x].size()<se[y].size()) swap(x,y);
		for (it=se[y].begin();it!=se[y].end();it++)
		{
			int to=(*it);
			ma[mp(to,y)]=ma[mp(y,to)]=0;
			se[to].erase(y);se[to].insert(x);
			ma[mp(to,x)]++;ma[mp(x,to)]++;
			if (ma[mp(to,x)]==2) que.push(mp(to,x));
			se[x].insert(to);
		}
		f[y]=x;
		se[y].clear();
	}
	int s=0;
	for (int i=1;i<=n;i++) s+=f[i]==i;
	if (s==1) puts("YES");
	else puts("NO");
	return 0;
}