#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<utility>
#define lint long long
#define P(x,y) ((lint)x*n+y)
#define mp make_pair
#define fir first
#define sec second
#define N 100010
using namespace std;
typedef pair<int,int> pii;
map<lint,int> f;
set<int> g[N];
queue<pii> q;
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=2*n-2;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		f[P(x,y)]++,f[P(y,x)]++;
		g[x].insert(y),g[y].insert(x);
		if(f[P(x,y)]==2) q.push(mp(x,y));
	}
	int ans=0;
	while(!q.empty())
	{
		int x=q.front().fir,y=q.front().sec;q.pop();
		if(f[P(x,y)]==2)
		{
			if(g[x].size()<g[y].size()) swap(x,y);
			while(!g[y].empty())
			{
				int t=*g[y].begin();
				f[P(x,t)]+=f[P(y,t)],f[P(y,t)]=0;
				f[P(t,x)]+=f[P(t,y)],f[P(t,y)]=0;
				g[x].insert(t),g[t].insert(x);
				g[y].erase(t),g[t].erase(y);
				if(f[P(x,t)]==2) q.push(mp(x,t));
			}
			ans++;
		}
	}
	if(ans==n-1) printf("YES\n");
	else printf("NO\n");return 0;
}
### 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
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<utility>
#define lint long long
#define P(x,y) ((lint)x*n+y)
#define mp make_pair
#define fir first
#define sec second
#define N 100010
using namespace std;
typedef pair<int,int> pii;
map<lint,int> f;
set<int> g[N];
queue<pii> q;
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=2*n-2;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		f[P(x,y)]++,f[P(y,x)]++;
		g[x].insert(y),g[y].insert(x);
		if(f[P(x,y)]==2) q.push(mp(x,y));
	}
	int ans=0;
	while(!q.empty())
	{
		int x=q.front().fir,y=q.front().sec;q.pop();
		if(f[P(x,y)]==2)
		{
			if(g[x].size()<g[y].size()) swap(x,y);
			while(!g[y].empty())
			{
				int t=*g[y].begin();
				f[P(x,t)]+=f[P(y,t)],f[P(y,t)]=0;
				f[P(t,x)]+=f[P(t,y)],f[P(t,y)]=0;
				g[x].insert(t),g[t].insert(x);
				g[y].erase(t),g[t].erase(y);
				if(f[P(x,t)]==2) q.push(mp(x,t));
			}
			ans++;
		}
	}
	if(ans==n-1) printf("YES\n");
	else printf("NO\n");return 0;
}
```