#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
map<ll, int> cnt;
set<int> g[100005];
vector<int> v1, v2;
int main() {
	scanf("%d", &n);
	for(int i=0; i<2*n-2; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		cnt[(ll)u*n+v]++;
		cnt[(ll)v*n+u]++;
		g[u].insert(v);
		g[v].insert(u);
		if(cnt[(ll)u*n+v]==2) {
			v1.push_back(u);
			v2.push_back(v);
		}
	}
	int ans=0;
	for(int i=0; i<v1.size(); i++) {
		if(cnt[(ll)v1[i]*n+v2[i]]==2) {
			ans++;
			if(g[v1[i]].size()<g[v2[i]].size())
				swap(v1[i], v2[i]);
			while(g[v2[i]].size()) {
				int w=*g[v2[i]].begin();
				cnt[(ll)v1[i]*n+w]+=cnt[(ll)v2[i]*n+w];
				cnt[(ll)w*n+v1[i]]+=cnt[(ll)w*n+v2[i]];
				cnt[(ll)v2[i]*n+w]=cnt[(ll)w*n+v2[i]]=0;
				g[v1[i]].insert(w);
				g[w].erase(v2[i]);
				g[w].insert(v1[i]);
				if(cnt[(ll)v1[i]*n+w]==2) {
					v1.push_back(v1[i]);
					v2.push_back(w);
				}
				g[v2[i]].erase(w);
			}
		}
	}
	if(ans==n-1)
		puts("YES");
	else
		puts("NO");
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;
typedef long long ll;
int n;
map<ll, int> cnt;
set<int> g[100005];
vector<int> v1, v2;
int main() {
	scanf("%d", &n);
	for(int i=0; i<2*n-2; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		cnt[(ll)u*n+v]++;
		cnt[(ll)v*n+u]++;
		g[u].insert(v);
		g[v].insert(u);
		if(cnt[(ll)u*n+v]==2) {
			v1.push_back(u);
			v2.push_back(v);
		}
	}
	int ans=0;
	for(int i=0; i<v1.size(); i++) {
		if(cnt[(ll)v1[i]*n+v2[i]]==2) {
			ans++;
			if(g[v1[i]].size()<g[v2[i]].size())
				swap(v1[i], v2[i]);
			while(g[v2[i]].size()) {
				int w=*g[v2[i]].begin();
				cnt[(ll)v1[i]*n+w]+=cnt[(ll)v2[i]*n+w];
				cnt[(ll)w*n+v1[i]]+=cnt[(ll)w*n+v2[i]];
				cnt[(ll)v2[i]*n+w]=cnt[(ll)w*n+v2[i]]=0;
				g[v1[i]].insert(w);
				g[w].erase(v2[i]);
				g[w].insert(v1[i]);
				if(cnt[(ll)v1[i]*n+w]==2) {
					v1.push_back(v1[i]);
					v2.push_back(w);
				}
				g[v2[i]].erase(w);
			}
		}
	}
	if(ans==n-1)
		puts("YES");
	else
		puts("NO");
	return 0;
}
```