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

const int N = (int) 1e5;
map<int, int> graph[N];
int lab[N];

int get(int u) {
	return lab[u] == u ? u : lab[u] = get(lab[u]);
}

void setRoot(int v) {
	if (lab[v] == v) return;
	int x = get(v);
	lab[x] = v;
	lab[v] = v;
}

void merge(int u, int v) {
	u = get(u);
	v = get(v);
	if (u != v) lab[u] = v;
}

int main() {
	ios::sync_with_stdio(false);
	int n; cin >> n;
	for (int i = 0; i < n; ++i) lab[i] = i;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v; --u; --v;
		graph[u][v] |= 1;
		graph[v][u] |= 1;
	}
	queue<pair<int, int> > q;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v; --u; --v;
// 		if (u > v) swap(u, v);
		graph[u][v] |= 2;
		graph[v][u] |= 2;
		if (graph[u][v] == 3) q.emplace(u, v);
	}
	int cnt = 0;
	while (!q.empty()) {
		int u = get(q.front().first);
		int v = get(q.front().second);
		q.pop();
// 		cout << u << " " << v << endl;
		if (u == v) continue;
		++cnt;
		if (graph[u].size() > graph[v].size()) swap(u, v);
		graph[u].erase(v);
		graph[v].erase(u);
		for (auto foo : graph[u]) {
			int w = foo.first;
			int val = foo.second;
			graph[w].erase(u);
			int tmp = graph[w][v] = graph[v][w] |= val;
			if (tmp == 3) {
				q.emplace(v, w);
			}
		}
		graph[u].clear();
// 		setRoot(v);
		merge(u, v);
	}
// 	for (int i = 0; i < n; ++i) cout << graph[i].size() << endl;
	cout << (cnt == n - 1 ? "YES" : "NO") << endl;
	return 0;
}

### Prompt

Please provide a cpp coded solution 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 <bits/stdc++.h>
using namespace std;

const int N = (int) 1e5;
map<int, int> graph[N];
int lab[N];

int get(int u) {
	return lab[u] == u ? u : lab[u] = get(lab[u]);
}

void setRoot(int v) {
	if (lab[v] == v) return;
	int x = get(v);
	lab[x] = v;
	lab[v] = v;
}

void merge(int u, int v) {
	u = get(u);
	v = get(v);
	if (u != v) lab[u] = v;
}

int main() {
	ios::sync_with_stdio(false);
	int n; cin >> n;
	for (int i = 0; i < n; ++i) lab[i] = i;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v; --u; --v;
		graph[u][v] |= 1;
		graph[v][u] |= 1;
	}
	queue<pair<int, int> > q;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v; --u; --v;
// 		if (u > v) swap(u, v);
		graph[u][v] |= 2;
		graph[v][u] |= 2;
		if (graph[u][v] == 3) q.emplace(u, v);
	}
	int cnt = 0;
	while (!q.empty()) {
		int u = get(q.front().first);
		int v = get(q.front().second);
		q.pop();
// 		cout << u << " " << v << endl;
		if (u == v) continue;
		++cnt;
		if (graph[u].size() > graph[v].size()) swap(u, v);
		graph[u].erase(v);
		graph[v].erase(u);
		for (auto foo : graph[u]) {
			int w = foo.first;
			int val = foo.second;
			graph[w].erase(u);
			int tmp = graph[w][v] = graph[v][w] |= val;
			if (tmp == 3) {
				q.emplace(v, w);
			}
		}
		graph[u].clear();
// 		setRoot(v);
		merge(u, v);
	}
// 	for (int i = 0; i < n; ++i) cout << graph[i].size() << endl;
	cout << (cnt == n - 1 ? "YES" : "NO") << endl;
	return 0;
}

```