#include <utility>
#include <cstdio>
#include <set>
int fa[100005];
std::set<int> adj[100005];
std::set<std::pair<int, int> > more;
int getfa(int u)
{
	if (u == fa[u])
		return u;
	return fa[u] = getfa(fa[u]);
}
inline void add_edge(int u, int v)
{
	if (adj[u].count(v))
	{
		adj[u].erase(v);
		if (u > v)
			std::swap(u, v);
		more.insert({u, v});
	}
	else
		adj[u].insert(v);
}
int main()
{
	// freopen("AGC014-E.in", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		fa[i] = i;
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(--u, --v);
		add_edge(v, u);
	}
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(--u, --v);
		add_edge(v, u);
	}
	int rem = n - 1;
	while (!more.empty())
	{
		auto e = *more.begin();
		more.erase(e);
		auto u = getfa(e.first), v = getfa(e.second);
		if (adj[u].size() > adj[v].size())
			std::swap(u, v);
		fa[u] = v;
		for (auto x : adj[u])
		{
			adj[x].erase(u);
			add_edge(v, x);
			add_edge(x, v);
		}
		adj[u].clear();
		rem--;
	}
	puts(!rem ? "YES" : "NO");
	return 0;
}

### Prompt

Develop a solution in Cpp 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 <utility>
#include <cstdio>
#include <set>
int fa[100005];
std::set<int> adj[100005];
std::set<std::pair<int, int> > more;
int getfa(int u)
{
	if (u == fa[u])
		return u;
	return fa[u] = getfa(fa[u]);
}
inline void add_edge(int u, int v)
{
	if (adj[u].count(v))
	{
		adj[u].erase(v);
		if (u > v)
			std::swap(u, v);
		more.insert({u, v});
	}
	else
		adj[u].insert(v);
}
int main()
{
	// freopen("AGC014-E.in", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		fa[i] = i;
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(--u, --v);
		add_edge(v, u);
	}
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(--u, --v);
		add_edge(v, u);
	}
	int rem = n - 1;
	while (!more.empty())
	{
		auto e = *more.begin();
		more.erase(e);
		auto u = getfa(e.first), v = getfa(e.second);
		if (adj[u].size() > adj[v].size())
			std::swap(u, v);
		fa[u] = v;
		for (auto x : adj[u])
		{
			adj[x].erase(u);
			add_edge(v, x);
			add_edge(x, v);
		}
		adj[u].clear();
		rem--;
	}
	puts(!rem ? "YES" : "NO");
	return 0;
}

```