#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100;
unordered_set<int> to[maxn];
int uf[maxn];
int find(int x){
	if(x == uf[x]) return x;
	return uf[x] = find(uf[x]);
}
using ii = pair<int,int>;
map<ii, int> c_edges;
queue<ii> q;
void upd_edge(int u, int v, int d){
	if(u == v) return;
	if(u > v) swap(u, v);
	ii e = {u, v};
	int cur_c = c_edges[e];
	cur_c += d;
	c_edges[e] = cur_c;
	if(cur_c == 2) q.push(e);
}
int join(int u, int v){
	u = find(u);
	v = find(v);
	if(u == v) return false;
	if(to[u].size() < to[v].size()) swap(u, v);
	to[u].erase(v);
	to[v].erase(u);
	for(int s : to[v]){
		upd_edge(v, s, -1);
		to[u].insert(s);
		to[s].erase(v);
		to[s].insert(u);
		upd_edge(u, s, 1);
	}
	to[v].clear();
	uf[v] = u;
	return true;
}
int main(){
	int n; cin >> n;
	iota(uf, uf + maxn, 0);
	for(int rot = 0; rot < 2; rot++){
		for(int i = 0; i < n - 1; i++){
			int u, v; cin >> u >> v;
			upd_edge(u, v, 1);
			to[u].insert(v);
			to[v].insert(u);
		}
	}
	for(int i = 0; i < n - 1; i++){
		if(q.empty()){
			cout << "NO" << endl;
			return 0;
		}
		int u, v;
		tie(u, v) = q.front(); q.pop();
		upd_edge(u, v, -1);
		upd_edge(u, v, -1);
		join(u, v);
	}
	cout << "YES" << endl;
	return 0;
}

### Prompt

Your task is to create 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 maxn = 100100;
unordered_set<int> to[maxn];
int uf[maxn];
int find(int x){
	if(x == uf[x]) return x;
	return uf[x] = find(uf[x]);
}
using ii = pair<int,int>;
map<ii, int> c_edges;
queue<ii> q;
void upd_edge(int u, int v, int d){
	if(u == v) return;
	if(u > v) swap(u, v);
	ii e = {u, v};
	int cur_c = c_edges[e];
	cur_c += d;
	c_edges[e] = cur_c;
	if(cur_c == 2) q.push(e);
}
int join(int u, int v){
	u = find(u);
	v = find(v);
	if(u == v) return false;
	if(to[u].size() < to[v].size()) swap(u, v);
	to[u].erase(v);
	to[v].erase(u);
	for(int s : to[v]){
		upd_edge(v, s, -1);
		to[u].insert(s);
		to[s].erase(v);
		to[s].insert(u);
		upd_edge(u, s, 1);
	}
	to[v].clear();
	uf[v] = u;
	return true;
}
int main(){
	int n; cin >> n;
	iota(uf, uf + maxn, 0);
	for(int rot = 0; rot < 2; rot++){
		for(int i = 0; i < n - 1; i++){
			int u, v; cin >> u >> v;
			upd_edge(u, v, 1);
			to[u].insert(v);
			to[v].insert(u);
		}
	}
	for(int i = 0; i < n - 1; i++){
		if(q.empty()){
			cout << "NO" << endl;
			return 0;
		}
		int u, v;
		tie(u, v) = q.front(); q.pop();
		upd_edge(u, v, -1);
		upd_edge(u, v, -1);
		join(u, v);
	}
	cout << "YES" << endl;
	return 0;
}

```