#include <bits/stdc++.h>

using namespace std;

#define M 200005
#define mp make_pair

int n, fa[M];

set<int> se[M];
map< pair<int, int>, int > ext;
queue< pair<int, int> > q;

int findf(int x)
{ return fa[x] == x ? x : fa[x] = findf(fa[x]); }

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '0') f = -1; ch = getchar(); }
	while (isdigit(ch)) { (x *= 10) += ch - '0'; ch = getchar(); }
	return x * f;
}

int main()
{
	n = read();
	for (int i = 1, u = 0, v = 0; i <= n * 2 - 2; i++)
	{
		u = read(), v = read();
		se[u].insert(v); se[v].insert(u);
		ext[mp(u, v)]++; ext[mp(v, u)]++;
		if (ext[mp(u, v)] == 2) q.push(mp(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();
		int uf = findf(u), vf = findf(v);
		se[uf].erase(vf), se[vf].erase(uf);
		if (se[uf].size() < se[vf].size()) swap(uf, vf);
		for (auto to : se[vf])
		{
			ext[mp(to, vf)] = ext[mp(vf, to)] = 0;
			se[to].erase(vf); se[to].insert(uf); se[uf].insert(to);
			ext[mp(to, uf)]++; ext[mp(uf, to)]++;
			if (ext[mp(to, uf)] == 2) q.push(mp(to, uf));
		}
		fa[vf] = uf; se[vf].clear();
	}

	int cnt = 0;
	for (int i = 1; i <= n; i++)
		if (fa[i] == i) cnt++;
	puts(cnt > 1 ? "NO" : "YES");

	return 0;
}
### Prompt

Your challenge is to write 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>

using namespace std;

#define M 200005
#define mp make_pair

int n, fa[M];

set<int> se[M];
map< pair<int, int>, int > ext;
queue< pair<int, int> > q;

int findf(int x)
{ return fa[x] == x ? x : fa[x] = findf(fa[x]); }

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '0') f = -1; ch = getchar(); }
	while (isdigit(ch)) { (x *= 10) += ch - '0'; ch = getchar(); }
	return x * f;
}

int main()
{
	n = read();
	for (int i = 1, u = 0, v = 0; i <= n * 2 - 2; i++)
	{
		u = read(), v = read();
		se[u].insert(v); se[v].insert(u);
		ext[mp(u, v)]++; ext[mp(v, u)]++;
		if (ext[mp(u, v)] == 2) q.push(mp(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();
		int uf = findf(u), vf = findf(v);
		se[uf].erase(vf), se[vf].erase(uf);
		if (se[uf].size() < se[vf].size()) swap(uf, vf);
		for (auto to : se[vf])
		{
			ext[mp(to, vf)] = ext[mp(vf, to)] = 0;
			se[to].erase(vf); se[to].insert(uf); se[uf].insert(to);
			ext[mp(to, uf)]++; ext[mp(uf, to)]++;
			if (ext[mp(to, uf)] == 2) q.push(mp(to, uf));
		}
		fa[vf] = uf; se[vf].clear();
	}

	int cnt = 0;
	for (int i = 1; i <= n; i++)
		if (fa[i] == i) cnt++;
	puts(cnt > 1 ? "NO" : "YES");

	return 0;
}
```