#include<iostream>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
using namespace std;
const int N=100001;
int n;
map<pair<int,int>,int> book;
set<int> e[N];
queue<pair<int,int> >q;
int f[N];
int find(int v)
{
	if(v==f[v]) return v;
	f[v]=find(f[v]);
	return f[v];
}
pair<int,int> make(int x,int y)
{
    if(x>y) swap(x,y);
    return make_pair(x,y);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    	f[i]=i;
    for(int i=1;i<=(n-1)*2;i++)
	{
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].insert(y);
		e[y].insert(x);
        pair<int,int> t=make(x,y);
        book[t]++;
        if(book[t]==2) q.push(t);
        if(book[t]>2)
        {
        	printf("NO");
        	return 0;
		}
    }
    for(int i=1;i<n;i++)
	{
        if(q.empty())
        {
        	printf("NO");
        	return 0;
		}
        int x=0,y=0;
        while(x==y)
        {
        	x=find(q.front().first),y=find(q.front().second);
        	q.pop();
		}
        if(e[x].size()>e[y].size()) swap(x,y);
        f[x]=y;
        for(set<int>::iterator it=e[x].begin();it!=e[x].end();it++)
		{
            int v=find(*it);
            e[v].erase(e[v].find(x));
            if(v==y) continue;
            e[v].insert(y),e[y].insert(v);
            pair<int,int> t=make(v,y);
            book[t]++;
            if(book[t]==2) q.push(t);
        }
        e[x].clear();
    }
    printf("YES");
    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<iostream>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
using namespace std;
const int N=100001;
int n;
map<pair<int,int>,int> book;
set<int> e[N];
queue<pair<int,int> >q;
int f[N];
int find(int v)
{
	if(v==f[v]) return v;
	f[v]=find(f[v]);
	return f[v];
}
pair<int,int> make(int x,int y)
{
    if(x>y) swap(x,y);
    return make_pair(x,y);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    	f[i]=i;
    for(int i=1;i<=(n-1)*2;i++)
	{
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].insert(y);
		e[y].insert(x);
        pair<int,int> t=make(x,y);
        book[t]++;
        if(book[t]==2) q.push(t);
        if(book[t]>2)
        {
        	printf("NO");
        	return 0;
		}
    }
    for(int i=1;i<n;i++)
	{
        if(q.empty())
        {
        	printf("NO");
        	return 0;
		}
        int x=0,y=0;
        while(x==y)
        {
        	x=find(q.front().first),y=find(q.front().second);
        	q.pop();
		}
        if(e[x].size()>e[y].size()) swap(x,y);
        f[x]=y;
        for(set<int>::iterator it=e[x].begin();it!=e[x].end();it++)
		{
            int v=find(*it);
            e[v].erase(e[v].find(x));
            if(v==y) continue;
            e[v].insert(y),e[y].insert(v);
            pair<int,int> t=make(v,y);
            book[t]++;
            if(book[t]==2) q.push(t);
        }
        e[x].clear();
    }
    printf("YES");
    return 0;
}
```