#include<bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 9;
 
set<int>r[N], b[N];
using pii = pair<int, int>;
queue<pii>Q;
int n, f[N];
int find (int x) { while (x != f[x]) x = f[x] = f[f[x]]; return x; }
 
int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) f[i] = i;
	for (int i = 1, u, v; i < n; ++i) scanf("%d%d", &u, &v), b[u].insert(v), b[v].insert(u);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v), r[u].insert(v), r[v].insert(u);
		if (b[u].count(v)) Q.push(pii(u, v)); 
	} 
	for (int i = 1, x, y; i < n; ++i) {
		if (Q.empty()) return puts("NO"), 0;
		x = find(Q.front().first); y = find(Q.front().second); Q.pop();
		if (x == y) { --i; continue ; }
		r[x].erase(y); r[y].erase(x); b[x].erase(y); b[y].erase(x);
		if (r[x].size() > r[y].size()) x ^= y ^= x ^= y;
		f[x] = y;
		for (auto v : r[x]) if (b[y].count(v)) Q.push(pii(v, y));
		for (auto v : b[x]) if (r[y].count(v)) Q.push(pii(v, y));
		for (auto v : r[x]) {
			r[y].insert(v), r[v].erase(x), r[v].insert(y);
		}
		for (auto v : b[x]) {
			b[y].insert(v), b[v].erase(x), b[v].insert(y);
		}
	}
	return puts("YES"), 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>
using namespace std;
 
const int N = 1e5 + 9;
 
set<int>r[N], b[N];
using pii = pair<int, int>;
queue<pii>Q;
int n, f[N];
int find (int x) { while (x != f[x]) x = f[x] = f[f[x]]; return x; }
 
int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) f[i] = i;
	for (int i = 1, u, v; i < n; ++i) scanf("%d%d", &u, &v), b[u].insert(v), b[v].insert(u);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v), r[u].insert(v), r[v].insert(u);
		if (b[u].count(v)) Q.push(pii(u, v)); 
	} 
	for (int i = 1, x, y; i < n; ++i) {
		if (Q.empty()) return puts("NO"), 0;
		x = find(Q.front().first); y = find(Q.front().second); Q.pop();
		if (x == y) { --i; continue ; }
		r[x].erase(y); r[y].erase(x); b[x].erase(y); b[y].erase(x);
		if (r[x].size() > r[y].size()) x ^= y ^= x ^= y;
		f[x] = y;
		for (auto v : r[x]) if (b[y].count(v)) Q.push(pii(v, y));
		for (auto v : b[x]) if (r[y].count(v)) Q.push(pii(v, y));
		for (auto v : r[x]) {
			r[y].insert(v), r[v].erase(x), r[v].insert(y);
		}
		for (auto v : b[x]) {
			b[y].insert(v), b[v].erase(x), b[v].insert(y);
		}
	}
	return puts("YES"), 0;
}
```