#include <cstdio>
#include <map>
#include <vector>
#define pb push_back
#define fs first
#define sc second
#define mp make_pair
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=1e5+5;
typedef pair<int,int> Pair;
Pair q[2*N];
int n,i,j,u,v,l,r,cnt,dgr,sz[N],fa[N];
map<Pair,int> g;
vector<int> E[N];
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
int main()
{
//	freopen("r&b.in","r",stdin);
//	freopen("r&b.out","w",stdout);
	scanf("%d",&n);
	rep(i,1,2*n-2)
	{
		scanf("%d%d",&u,&v);
		E[u].pb(v); E[v].pb(u);
		g[mp(u,v)]++; g[mp(v,u)]++; sz[u]++; sz[v]++;
		if (g[mp(u,v)]==2) q[++r]=mp(u,v);
	}
	rep(i,1,n) fa[i]=i;
	rep(l,1,r)
	{
		u=find(q[l].fs); v=find(q[l].sc);
		if (u==v) continue;
		cnt++;
		if (sz[u]<sz[v]) swap(u,v);
		fa[v]=u;
		rep(i,0,sz[v]-1)
		{
			j=find(E[v][i]);
			if (j==u) continue;
			E[u].pb(j);
			sz[u]++;
			g[mp(u,j)]++; g[mp(j,u)]++;
			
			dgr=g[mp(u,j)];
			if (dgr>2) {
				printf("NO\n");
				return 0;
			}
			else if (dgr==2) q[++r]=mp(u,j);
		}
	}
	if (cnt<n-1) printf("NO\n");
	else printf("YES\n");
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdio>
#include <map>
#include <vector>
#define pb push_back
#define fs first
#define sc second
#define mp make_pair
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=1e5+5;
typedef pair<int,int> Pair;
Pair q[2*N];
int n,i,j,u,v,l,r,cnt,dgr,sz[N],fa[N];
map<Pair,int> g;
vector<int> E[N];
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
int main()
{
//	freopen("r&b.in","r",stdin);
//	freopen("r&b.out","w",stdout);
	scanf("%d",&n);
	rep(i,1,2*n-2)
	{
		scanf("%d%d",&u,&v);
		E[u].pb(v); E[v].pb(u);
		g[mp(u,v)]++; g[mp(v,u)]++; sz[u]++; sz[v]++;
		if (g[mp(u,v)]==2) q[++r]=mp(u,v);
	}
	rep(i,1,n) fa[i]=i;
	rep(l,1,r)
	{
		u=find(q[l].fs); v=find(q[l].sc);
		if (u==v) continue;
		cnt++;
		if (sz[u]<sz[v]) swap(u,v);
		fa[v]=u;
		rep(i,0,sz[v]-1)
		{
			j=find(E[v][i]);
			if (j==u) continue;
			E[u].pb(j);
			sz[u]++;
			g[mp(u,j)]++; g[mp(j,u)]++;
			
			dgr=g[mp(u,j)];
			if (dgr>2) {
				printf("NO\n");
				return 0;
			}
			else if (dgr==2) q[++r]=mp(u,j);
		}
	}
	if (cnt<n-1) printf("NO\n");
	else printf("YES\n");
	return 0;
}
```