#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cassert>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<utility>
#define pii pair<int,int> 
#define mkpr make_pair
using namespace std;

const int N = 1e5;
multiset<int> ae[N+3];
map<pii,int> g;
queue<pii> que;
int n;

void insertedge(int u,int v)
{
	if(u==v) return;
	if(u>v) swap(u,v);
	ae[u].insert(v); ae[v].insert(u);
	g[mkpr(u,v)]++;
	if(g[mkpr(u,v)]==2) {que.push(mkpr(u,v));}
}

void deleteedge(int u,int v)
{
	ae[v].erase(ae[v].find(u));
	if(u>v) swap(u,v);
	if(g.count(mkpr(u,v))) {g.erase(mkpr(u,v));}
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=2*(n-1); i++)
	{
		int u,v; scanf("%d%d",&u,&v);
		insertedge(u,v);
	}
	for(int i=1; i<n; i++)
	{
		while(1)
		{
			if(que.empty()) {puts("NO"); return 0;}
			int u = que.front().first,v = que.front().second; que.pop();
			if(!g.count(mkpr(u,v))) {continue;}
			if(ae[u].size()<ae[v].size()) swap(u,v);
			for(set<int>::iterator i=ae[v].begin(); i!=ae[v].end(); i++)
			{
				int x = *i;
				deleteedge(v,x);
				insertedge(u,x);
			}
			break;
		}
	}
	puts("YES");
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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<cstdlib>
#include<iostream>
#include<cassert>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<utility>
#define pii pair<int,int> 
#define mkpr make_pair
using namespace std;

const int N = 1e5;
multiset<int> ae[N+3];
map<pii,int> g;
queue<pii> que;
int n;

void insertedge(int u,int v)
{
	if(u==v) return;
	if(u>v) swap(u,v);
	ae[u].insert(v); ae[v].insert(u);
	g[mkpr(u,v)]++;
	if(g[mkpr(u,v)]==2) {que.push(mkpr(u,v));}
}

void deleteedge(int u,int v)
{
	ae[v].erase(ae[v].find(u));
	if(u>v) swap(u,v);
	if(g.count(mkpr(u,v))) {g.erase(mkpr(u,v));}
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=2*(n-1); i++)
	{
		int u,v; scanf("%d%d",&u,&v);
		insertedge(u,v);
	}
	for(int i=1; i<n; i++)
	{
		while(1)
		{
			if(que.empty()) {puts("NO"); return 0;}
			int u = que.front().first,v = que.front().second; que.pop();
			if(!g.count(mkpr(u,v))) {continue;}
			if(ae[u].size()<ae[v].size()) swap(u,v);
			for(set<int>::iterator i=ae[v].begin(); i!=ae[v].end(); i++)
			{
				int x = *i;
				deleteedge(v,x);
				insertedge(u,x);
			}
			break;
		}
	}
	puts("YES");
	return 0;
}
```