#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define mkp(a,b) make_pair(a, b)
#define pi pair < int , int >
#define N 100005

using namespace std;

int n;
int fa[N];

set < int > s[N];
map < pi, int> p;
queue < pi > Q;

int find(int);
int read();

int main(){
	n = read();
	for(int i = 3; i <= (n << 1); ++i){
		int u = read(), v = read();
		++p[mkp(u, v)];++p[mkp(v, u)];
		s[u].insert(v);s[v].insert(u);
		if(p[mkp(u, v)] == 2) Q.push(mkp(u, v));
	}
	for(int i = 1; i <= n; ++i)
		fa[i] = i;
	while(!Q.empty()){
		int u = Q.front().first, v = Q.front().second; Q.pop();
		u = find(u), v = find(v);
		s[u].erase(v);s[v].erase(u);
		if(s[u].size() > s[v].size()) swap(u, v);
		set < int > :: iterator it;
		it = s[u].begin();v = find(v);
		while(it != s[u].end()){
			int to = *it;
			p[mkp(to, u)] = p[mkp(u, to)] = 0;
			s[v].insert(to);s[to].erase(u);s[to].insert(v);
			++p[mkp(v, to)];++p[mkp(to, v)];
			if(p[mkp(v, to)] == 2) Q.push(mkp(to, v));
			++it;
		}
		s[u].clear();fa[u] = v;
	}
	int cnt = 0;
	for(int i = 1; i <= n; ++i) cnt += (fa[i] == i);
	if(cnt == 1) puts("YES");
	else puts("NO");
	return 0;
}

int read(){
	int x = 0;char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return x;
}

int find(int x){
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}
### Prompt

Your task is to create 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
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define mkp(a,b) make_pair(a, b)
#define pi pair < int , int >
#define N 100005

using namespace std;

int n;
int fa[N];

set < int > s[N];
map < pi, int> p;
queue < pi > Q;

int find(int);
int read();

int main(){
	n = read();
	for(int i = 3; i <= (n << 1); ++i){
		int u = read(), v = read();
		++p[mkp(u, v)];++p[mkp(v, u)];
		s[u].insert(v);s[v].insert(u);
		if(p[mkp(u, v)] == 2) Q.push(mkp(u, v));
	}
	for(int i = 1; i <= n; ++i)
		fa[i] = i;
	while(!Q.empty()){
		int u = Q.front().first, v = Q.front().second; Q.pop();
		u = find(u), v = find(v);
		s[u].erase(v);s[v].erase(u);
		if(s[u].size() > s[v].size()) swap(u, v);
		set < int > :: iterator it;
		it = s[u].begin();v = find(v);
		while(it != s[u].end()){
			int to = *it;
			p[mkp(to, u)] = p[mkp(u, to)] = 0;
			s[v].insert(to);s[to].erase(u);s[to].insert(v);
			++p[mkp(v, to)];++p[mkp(to, v)];
			if(p[mkp(v, to)] == 2) Q.push(mkp(to, v));
			++it;
		}
		s[u].clear();fa[u] = v;
	}
	int cnt = 0;
	for(int i = 1; i <= n; ++i) cnt += (fa[i] == i);
	if(cnt == 1) puts("YES");
	else puts("NO");
	return 0;
}

int read(){
	int x = 0;char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return x;
}

int find(int x){
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}
```