//Heaplax
//别让自己后悔
#include<bits/stdc++.h>
#define N 100001
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
queue<pair<int,int>> que;
int n;
map<int,int> edge[N],test;
int main()
{
	re(n);
	for(int i=1,u,v;i<n;++i)
	{
		re(u),re(v);
		++edge[u][v];
		++edge[v][u];
	}
	for(int i=1,u,v;i<n;++i)
	{
		re(u),re(v);
		++edge[u][v];
		++edge[v][u];
		if(edge[u][v]==2)
			que.push(make_pair(u,v));
	}
	while(!que.empty())
	{
		int u=que.front().first,v=que.front().second;
		que.pop();
		if(edge[u][v]!=2)continue;
		--n;
		if(edge[u].size() < edge[v].size())swap(u,v);
		edge[u].erase(v);
		edge[v].erase(u);
		for(auto i:edge[v])
		{
			edge[u][i.first]+=i.second;
			edge[i.first][u]+=i.second;
			if(edge[u][i.first]==2)
				que.push(make_pair(u,i.first));
			edge[i.first].erase(v);
		}
		edge[v].clear();
	}
	puts(n==1?"YES":"NO");
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
//Heaplax
//别让自己后悔
#include<bits/stdc++.h>
#define N 100001
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
queue<pair<int,int>> que;
int n;
map<int,int> edge[N],test;
int main()
{
	re(n);
	for(int i=1,u,v;i<n;++i)
	{
		re(u),re(v);
		++edge[u][v];
		++edge[v][u];
	}
	for(int i=1,u,v;i<n;++i)
	{
		re(u),re(v);
		++edge[u][v];
		++edge[v][u];
		if(edge[u][v]==2)
			que.push(make_pair(u,v));
	}
	while(!que.empty())
	{
		int u=que.front().first,v=que.front().second;
		que.pop();
		if(edge[u][v]!=2)continue;
		--n;
		if(edge[u].size() < edge[v].size())swap(u,v);
		edge[u].erase(v);
		edge[v].erase(u);
		for(auto i:edge[v])
		{
			edge[u][i.first]+=i.second;
			edge[i.first][u]+=i.second;
			if(edge[u][i.first]==2)
				que.push(make_pair(u,i.first));
			edge[i.first].erase(v);
		}
		edge[v].clear();
	}
	puts(n==1?"YES":"NO");
}

```