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

int n;
set<pair<int, bool> > edges[100001];

int main() {
	stack<pair<int, int> > stk;
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int x, y;
		scanf("%d%d", &x, &y), --x, --y;
		edges[x].insert(make_pair(y, 0));
		edges[y].insert(make_pair(x, 0));
	}
	for (int i = 1; i < n; ++i) {
		int x, y;
		scanf("%d%d", &x, &y), --x, --y;
		edges[x].insert(make_pair(y, 1));
		edges[y].insert(make_pair(x, 1));
		stk.push(make_pair(x, y));
	}
	int cnt = 0;
	while (!stk.empty()) {
		pair<int, int> edge = stk.top();
		stk.pop();
		int x = edge.first, y = edge.second;
		auto it = edges[x].find(make_pair(y, 0));
		if (it != edges[x].end()) {
			__typeof(it) it2 = it;
			++it2;
			if (it2 != edges[x].end() && it2->first == y) {
				if (edges[x].size() < edges[y].size()) swap(x, y);
				++cnt;
				edges[x].erase(make_pair(y, 0));
				edges[x].erase(make_pair(y, 1));
				edges[y].erase(make_pair(x, 0));
				edges[y].erase(make_pair(x, 1));
				for (auto it3 = edges[y].begin(); it3 != edges[y].end();) {
					stk.push(make_pair(x, it3->first));
					edges[it3->first].erase(make_pair(y, it3->second));
					edges[it3->first].insert(make_pair(x, it3->second));
					edges[x].insert(*it3);
					it3 = edges[y].erase(it3);
				}
			}
		}
	}
	puts(cnt == n - 1 ? "YES" : "NO");
}
### 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;

int n;
set<pair<int, bool> > edges[100001];

int main() {
	stack<pair<int, int> > stk;
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int x, y;
		scanf("%d%d", &x, &y), --x, --y;
		edges[x].insert(make_pair(y, 0));
		edges[y].insert(make_pair(x, 0));
	}
	for (int i = 1; i < n; ++i) {
		int x, y;
		scanf("%d%d", &x, &y), --x, --y;
		edges[x].insert(make_pair(y, 1));
		edges[y].insert(make_pair(x, 1));
		stk.push(make_pair(x, y));
	}
	int cnt = 0;
	while (!stk.empty()) {
		pair<int, int> edge = stk.top();
		stk.pop();
		int x = edge.first, y = edge.second;
		auto it = edges[x].find(make_pair(y, 0));
		if (it != edges[x].end()) {
			__typeof(it) it2 = it;
			++it2;
			if (it2 != edges[x].end() && it2->first == y) {
				if (edges[x].size() < edges[y].size()) swap(x, y);
				++cnt;
				edges[x].erase(make_pair(y, 0));
				edges[x].erase(make_pair(y, 1));
				edges[y].erase(make_pair(x, 0));
				edges[y].erase(make_pair(x, 1));
				for (auto it3 = edges[y].begin(); it3 != edges[y].end();) {
					stk.push(make_pair(x, it3->first));
					edges[it3->first].erase(make_pair(y, it3->second));
					edges[it3->first].insert(make_pair(x, it3->second));
					edges[x].insert(*it3);
					it3 = edges[y].erase(it3);
				}
			}
		}
	}
	puts(cnt == n - 1 ? "YES" : "NO");
}
```