#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <map>
#include <set>
using namespace std;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 100000;

int n;
multiset<int> adj[MaxN + 1];

map<pair<int, int>, int> all;

queue<pair<int, int>> dup;

inline void add(int u, int v)
{
	if (u > v)
		swap(u, v);

	adj[u].insert(v);
	adj[v].insert(u);

	pair<int, int> e(u, v);
	if (++all[e] > 1)
		dup.push(e);
}

int ufs[MaxN + 1];

int ufs_find(const int &u)
{
	return ufs[u] == u ? u : ufs[u] = ufs_find(ufs[u]);
}

int main()
{
	cin >> n;
	for (int i = 2; i < 2 * n; ++i)
	{
		int u = getint(), v = getint();
	//	adj[u].insert(v);
	//	adj[v].insert(u);

		add(u, v);
	}

	for (int u = 1; u <= n; ++u)
		ufs[u] = u;

	for (int k = 1; k < n; ++k)
	{
		int u = 0, v = 0;
		while (u == v && !dup.empty())
		{
			u = ufs_find(dup.front().first);
			v = ufs_find(dup.front().second);
			dup.pop();
		}

		if (u == v && dup.empty())
		{
			cout << "NO" << endl;
			return 0;
		}

		if (adj[u].size() < adj[v].size())
			swap(u, v);

		ufs[v] = u;

		for (auto i : adj[v])
		{
			int r = ufs_find(i);
			if (r == u)
				continue;

			adj[r].erase(adj[r].find(v));

			add(u, r);
		}

		adj[u].erase(v);
		adj[v].clear();
	}

	cout << "YES" << endl;

	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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <map>
#include <set>
using namespace std;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 100000;

int n;
multiset<int> adj[MaxN + 1];

map<pair<int, int>, int> all;

queue<pair<int, int>> dup;

inline void add(int u, int v)
{
	if (u > v)
		swap(u, v);

	adj[u].insert(v);
	adj[v].insert(u);

	pair<int, int> e(u, v);
	if (++all[e] > 1)
		dup.push(e);
}

int ufs[MaxN + 1];

int ufs_find(const int &u)
{
	return ufs[u] == u ? u : ufs[u] = ufs_find(ufs[u]);
}

int main()
{
	cin >> n;
	for (int i = 2; i < 2 * n; ++i)
	{
		int u = getint(), v = getint();
	//	adj[u].insert(v);
	//	adj[v].insert(u);

		add(u, v);
	}

	for (int u = 1; u <= n; ++u)
		ufs[u] = u;

	for (int k = 1; k < n; ++k)
	{
		int u = 0, v = 0;
		while (u == v && !dup.empty())
		{
			u = ufs_find(dup.front().first);
			v = ufs_find(dup.front().second);
			dup.pop();
		}

		if (u == v && dup.empty())
		{
			cout << "NO" << endl;
			return 0;
		}

		if (adj[u].size() < adj[v].size())
			swap(u, v);

		ufs[v] = u;

		for (auto i : adj[v])
		{
			int r = ufs_find(i);
			if (r == u)
				continue;

			adj[r].erase(adj[r].find(v));

			add(u, r);
		}

		adj[u].erase(v);
		adj[v].clear();
	}

	cout << "YES" << endl;

	return 0;
}
```