#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1.1e5;
int N;
set<int> adj[MAXN];

int par[MAXN];
int getpar(int a) {
	return par[a] == -1 ? a : (par[a] = getpar(par[a]));
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> N;
	for (int i = 0; i < N-1; i++) {
		int a, b; cin >> a >> b;
		adj[a].insert(b);
		adj[b].insert(a);
	}
	vector<pair<int, int>> todos;
	for (int i = 0; i < N-1; i++) {
		int a, b; cin >> a >> b;
		if (adj[a].count(b)) {
			todos.emplace_back(a, b);
		} else {
			adj[a].insert(b);
			adj[b].insert(a);
		}
	}

	for (int i = 1; i <= N; i++) par[i] = -1;
	int numEdges = 0;
	while (!todos.empty()) {
		int a, b; tie(a, b) = todos.back();
		todos.pop_back();
		a = getpar(a);
		b = getpar(b);
		if (a == b) continue;
		numEdges++;

		if (adj[a].size() > adj[b].size()) {
			swap(a, b);
		}
		adj[a].erase(b);
		adj[b].erase(a);
		par[a] = b;
		for (int c : adj[a]) {
			adj[c].erase(a);
			if (adj[b].count(c)) {
				todos.emplace_back(b, c);
			} else {
				adj[b].insert(c);
				adj[c].insert(b);
			}
		}
		set<int>().swap(adj[a]);
	}
	cout << (numEdges == N-1 ? "YES" : "NO") << '\n';

	return 0;
}

### Prompt

In cpp, your task is to solve 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;

const int MAXN = 1.1e5;
int N;
set<int> adj[MAXN];

int par[MAXN];
int getpar(int a) {
	return par[a] == -1 ? a : (par[a] = getpar(par[a]));
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> N;
	for (int i = 0; i < N-1; i++) {
		int a, b; cin >> a >> b;
		adj[a].insert(b);
		adj[b].insert(a);
	}
	vector<pair<int, int>> todos;
	for (int i = 0; i < N-1; i++) {
		int a, b; cin >> a >> b;
		if (adj[a].count(b)) {
			todos.emplace_back(a, b);
		} else {
			adj[a].insert(b);
			adj[b].insert(a);
		}
	}

	for (int i = 1; i <= N; i++) par[i] = -1;
	int numEdges = 0;
	while (!todos.empty()) {
		int a, b; tie(a, b) = todos.back();
		todos.pop_back();
		a = getpar(a);
		b = getpar(b);
		if (a == b) continue;
		numEdges++;

		if (adj[a].size() > adj[b].size()) {
			swap(a, b);
		}
		adj[a].erase(b);
		adj[b].erase(a);
		par[a] = b;
		for (int c : adj[a]) {
			adj[c].erase(a);
			if (adj[b].count(c)) {
				todos.emplace_back(b, c);
			} else {
				adj[b].insert(c);
				adj[c].insert(b);
			}
		}
		set<int>().swap(adj[a]);
	}
	cout << (numEdges == N-1 ? "YES" : "NO") << '\n';

	return 0;
}

```