#include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int N=100005;
typedef pair<int,int> PR;
int n,f[N];
multiset<int> S[N];
map<PR,int> mp;
queue<PR> Q;
int find(int x) {return x==f[x]?x:f[x]=find(f[x]);}
int main()
{
	int x,y;
	n=read();
	for(RI i=1;i<=n;++i) f[i]=i;
	for(RI i=1;i<=2*n-2;++i) {
		x=read(),y=read();
		S[x].insert(y),S[y].insert(x);
		++mp[(PR){x,y}],++mp[(PR){y,x}];
		if((mp[(PR){x,y}])==2) Q.push((PR){x,y});
	}
	multiset<int>::iterator it;
	while(!Q.empty()) {
		x=Q.front().first,y=Q.front().second,Q.pop();
		x=find(x),y=find(y);
		if(x==y) continue;
		S[x].erase(y),S[y].erase(x);
		if(S[x].size()<S[y].size()) swap(x,y);
		for(it=S[y].begin();it!=S[y].end();++it) {
			int z=*it;
			S[z].erase(y),S[z].insert(x);
			++mp[(PR){z,x}],++mp[(PR){x,z}];
			if(mp[(PR){z,x}]==2) Q.push((PR){z,x});
			S[x].insert(z);
		}
		f[y]=x,S[y].clear();
	}
	for(RI i=1;i<=n;++i) if(find(i)!=find(1)) {puts("NO");return 0;}
	puts("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<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int N=100005;
typedef pair<int,int> PR;
int n,f[N];
multiset<int> S[N];
map<PR,int> mp;
queue<PR> Q;
int find(int x) {return x==f[x]?x:f[x]=find(f[x]);}
int main()
{
	int x,y;
	n=read();
	for(RI i=1;i<=n;++i) f[i]=i;
	for(RI i=1;i<=2*n-2;++i) {
		x=read(),y=read();
		S[x].insert(y),S[y].insert(x);
		++mp[(PR){x,y}],++mp[(PR){y,x}];
		if((mp[(PR){x,y}])==2) Q.push((PR){x,y});
	}
	multiset<int>::iterator it;
	while(!Q.empty()) {
		x=Q.front().first,y=Q.front().second,Q.pop();
		x=find(x),y=find(y);
		if(x==y) continue;
		S[x].erase(y),S[y].erase(x);
		if(S[x].size()<S[y].size()) swap(x,y);
		for(it=S[y].begin();it!=S[y].end();++it) {
			int z=*it;
			S[z].erase(y),S[z].insert(x);
			++mp[(PR){z,x}],++mp[(PR){x,z}];
			if(mp[(PR){z,x}]==2) Q.push((PR){z,x});
			S[x].insert(z);
		}
		f[y]=x,S[y].clear();
	}
	for(RI i=1;i<=n;++i) if(find(i)!=find(1)) {puts("NO");return 0;}
	puts("YES");
	return 0;
}
```