//Love and Freedom.
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#define inf 20021225
#define ll long long
#define N 100100
#define pa pair<int,int>
#define fs first
#define se second
#define mp make_pair
using namespace std;
int read()
{
	int f=1,s=0; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9')	s=s*10+ch-'0',ch=getchar();
	return f*s;
}
map<pa,int> g;
multiset<int> ed[N];
queue<pa> nd;
void add(int x,int y)
{
	if(x==y) return; if(x>y) swap(x,y); ed[x].insert(y); ed[y].insert(x);
	g[mp(x,y)]++; if(g[mp(x,y)]==2)	nd.push(mp(x,y));
}
void del(int x,int y)
{
	ed[y].erase(ed[y].find(x)); if(x>y)	swap(x,y);
	if(g.count(mp(x,y))) g.erase(mp(x,y));
}
int main()
{
	int n=read();
	for(int i=1;i<=2*(n-1);i++)	add(read(),read());
	for(int i=1;i<n;i++)
	{
		while(1)
		{
			if(nd.empty())	return puts("NO"),0;
			int x=nd.front().fs,y=nd.front().se; nd.pop();
			if(!g.count(mp(x,y)))	continue;
			if(ed[x].size()<ed[y].size())	swap(x,y);
			for(multiset<int>::iterator it=ed[y].begin();it!=ed[y].end();it++)
			{
				int to=*it; del(y,to); add(x,to);
			}
			break;
		}
	}
	puts("YES");
	return 0;
}
### Prompt

Please formulate 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
//Love and Freedom.
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#define inf 20021225
#define ll long long
#define N 100100
#define pa pair<int,int>
#define fs first
#define se second
#define mp make_pair
using namespace std;
int read()
{
	int f=1,s=0; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9')	s=s*10+ch-'0',ch=getchar();
	return f*s;
}
map<pa,int> g;
multiset<int> ed[N];
queue<pa> nd;
void add(int x,int y)
{
	if(x==y) return; if(x>y) swap(x,y); ed[x].insert(y); ed[y].insert(x);
	g[mp(x,y)]++; if(g[mp(x,y)]==2)	nd.push(mp(x,y));
}
void del(int x,int y)
{
	ed[y].erase(ed[y].find(x)); if(x>y)	swap(x,y);
	if(g.count(mp(x,y))) g.erase(mp(x,y));
}
int main()
{
	int n=read();
	for(int i=1;i<=2*(n-1);i++)	add(read(),read());
	for(int i=1;i<n;i++)
	{
		while(1)
		{
			if(nd.empty())	return puts("NO"),0;
			int x=nd.front().fs,y=nd.front().se; nd.pop();
			if(!g.count(mp(x,y)))	continue;
			if(ed[x].size()<ed[y].size())	swap(x,y);
			for(multiset<int>::iterator it=ed[y].begin();it!=ed[y].end();it++)
			{
				int to=*it; del(y,to); add(x,to);
			}
			break;
		}
	}
	puts("YES");
	return 0;
}
```