#include <bits/stdc++.h>
#define N 100005

using namespace std;

int n, totBlock;
int size[N];
set<int> g[N];
queue<pair<int, int> > que;

namespace UFS {

int fa[N];
inline void init() { for(int i = 1; i <= n; ++i) fa[i] = i; }
inline int find(int u) { return fa[u] == u ? u : fa[u] = find(fa[u]); }

}

inline void read(int &x) {
	x = 0; char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
}

void work() {
	int u, v;
	pair<int, int> o;
	while(!que.empty()) {
		o = que.front(), que.pop();
		u = UFS :: find(o.first), v = UFS :: find(o.second);
		if(u != v) {
			--totBlock;
			if(size[u] > size[v]) swap(u, v);
			// merge
			UFS :: fa[u] = v;
			for(set<int> :: iterator it = g[u].begin(); it != g[u].end(); ++it) {
				int to = UFS :: find(*it);
				if(to == v) continue;
				if(g[to].find(v) != g[to].end()) {
					g[to].erase(u), --size[to];
					//g[v].erase(to), --size[v];
					que.push(make_pair(u, to));
				} else {
					g[to].erase(u), g[to].insert(v);
					g[v].insert(to), ++size[v];
				}
			}
		}
	}
}

int main() {
	read(n);
	UFS :: init();
	for(int i = 1, u, v; i <= 2 * n - 2; ++i) {
		read(u), read(v);
		if(g[u].find(v) != g[u].end()) {
			g[u].erase(v), --size[u];
			g[v].erase(u), --size[v];
			que.push(make_pair(u, v));
		} else {
			g[u].insert(v), ++size[u];
			g[v].insert(u), ++size[v];
		}
	}
	
	totBlock = n;
	work();
	
	puts(totBlock == 1 ? "YES" : "NO");
	return 0;
}
### 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 <bits/stdc++.h>
#define N 100005

using namespace std;

int n, totBlock;
int size[N];
set<int> g[N];
queue<pair<int, int> > que;

namespace UFS {

int fa[N];
inline void init() { for(int i = 1; i <= n; ++i) fa[i] = i; }
inline int find(int u) { return fa[u] == u ? u : fa[u] = find(fa[u]); }

}

inline void read(int &x) {
	x = 0; char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
}

void work() {
	int u, v;
	pair<int, int> o;
	while(!que.empty()) {
		o = que.front(), que.pop();
		u = UFS :: find(o.first), v = UFS :: find(o.second);
		if(u != v) {
			--totBlock;
			if(size[u] > size[v]) swap(u, v);
			// merge
			UFS :: fa[u] = v;
			for(set<int> :: iterator it = g[u].begin(); it != g[u].end(); ++it) {
				int to = UFS :: find(*it);
				if(to == v) continue;
				if(g[to].find(v) != g[to].end()) {
					g[to].erase(u), --size[to];
					//g[v].erase(to), --size[v];
					que.push(make_pair(u, to));
				} else {
					g[to].erase(u), g[to].insert(v);
					g[v].insert(to), ++size[v];
				}
			}
		}
	}
}

int main() {
	read(n);
	UFS :: init();
	for(int i = 1, u, v; i <= 2 * n - 2; ++i) {
		read(u), read(v);
		if(g[u].find(v) != g[u].end()) {
			g[u].erase(v), --size[u];
			g[v].erase(u), --size[v];
			que.push(make_pair(u, v));
		} else {
			g[u].insert(v), ++size[u];
			g[v].insert(u), ++size[v];
		}
	}
	
	totBlock = n;
	work();
	
	puts(totBlock == 1 ? "YES" : "NO");
	return 0;
}
```