#include <bits/stdc++.h>

const int MAXN = 200010;
typedef std::pair<int, int> pi;
std::set<int> qry[MAXN], out[MAXN];
std::set<pi> S;
int n;
void chk(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	if (out[x].count(y)) S.emplace(x, y);
}
void chk2(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	if (qry[x].count(y)) S.emplace(x, y);
}
void del(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	S.erase(pi(x, y));
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, t1, t2; i < n; ++i)
		std::cin >> t1 >> t2, out[t1].insert(t2), out[t2].insert(t1);
	for (int i = 1, t1, t2; i < n; ++i) {
		std::cin >> t1 >> t2;
		qry[t1].insert(t2), qry[t2].insert(t1);
		chk(t1, t2);
	}
	int cnt = 0;
	while (!S.empty()) {
		pi t = *S.begin(); S.erase(S.begin()); ++cnt;
		int x = t.first, y = t.second;
		if (out[x].size() + qry[x].size() > out[y].size() + qry[y].size())
			std::swap(x, y);
		for (auto u : out[x]) {
			out[u].erase(x), del(u, x);
			if (u != y) out[u].insert(y), out[y].insert(u);
		}
		for (auto u : qry[x]) {
			qry[u].erase(x), del(u, x);
			if (u != y) qry[u].insert(y), qry[y].insert(u);
		}
		for (auto u : out[x]) chk2(u, y);
		for (auto u : qry[x]) chk(u, y);
		out[x].clear(), qry[x].clear();
	}
	std::cout << (cnt == n - 1 ? "YES" : "NO") << std::endl;
	return 0;
}

### Prompt

Generate 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>

const int MAXN = 200010;
typedef std::pair<int, int> pi;
std::set<int> qry[MAXN], out[MAXN];
std::set<pi> S;
int n;
void chk(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	if (out[x].count(y)) S.emplace(x, y);
}
void chk2(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	if (qry[x].count(y)) S.emplace(x, y);
}
void del(int x, int y) {
	if (x > y) std::swap(x, y); if (x == y) return ;
	S.erase(pi(x, y));
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, t1, t2; i < n; ++i)
		std::cin >> t1 >> t2, out[t1].insert(t2), out[t2].insert(t1);
	for (int i = 1, t1, t2; i < n; ++i) {
		std::cin >> t1 >> t2;
		qry[t1].insert(t2), qry[t2].insert(t1);
		chk(t1, t2);
	}
	int cnt = 0;
	while (!S.empty()) {
		pi t = *S.begin(); S.erase(S.begin()); ++cnt;
		int x = t.first, y = t.second;
		if (out[x].size() + qry[x].size() > out[y].size() + qry[y].size())
			std::swap(x, y);
		for (auto u : out[x]) {
			out[u].erase(x), del(u, x);
			if (u != y) out[u].insert(y), out[y].insert(u);
		}
		for (auto u : qry[x]) {
			qry[u].erase(x), del(u, x);
			if (u != y) qry[u].insert(y), qry[y].insert(u);
		}
		for (auto u : out[x]) chk2(u, y);
		for (auto u : qry[x]) chk(u, y);
		out[x].clear(), qry[x].clear();
	}
	std::cout << (cnt == n - 1 ? "YES" : "NO") << std::endl;
	return 0;
}

```