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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

const int N = 100010;

std::set <int> set[N];
std::queue <std::pair <int, int>> q;
int n;
int fa[N], sz[N];

int find(int u){
	return fa[u] == u ? u : fa[u] = find(fa[u]);
}

void unite(int u, int v){
	int _u = find(u), _v = find(v);
	if (_u == _v){
		return;
	}
	fa[_u] = fa[_v];
	sz[_v] += sz[_u];
	find(u);
}

int main(){
	for (int i = 0; i < N; ++ i){
		fa[i] = i;
		sz[i] = 1;
	}
	scanf("%d", &n);
	for (int i = 0, u, v; i < n - 1 << 1; ++ i){
		scanf("%d%d", &u, &v);
		if (set[u].count(v)){
			q.push({u, v});
		}
		else{
			set[u].insert(v);
			set[v].insert(u);
		}
	}
	while (!q.empty()){
		std::pair <int, int> p = q.front();
		q.pop();
		int u = find(p.first), v = find(p.second);
		if (sz[u] > sz[v]){
			std::swap(u, v);
		}
		unite(u, v);
		set[u].erase(v);
		set[v].erase(u);
		for (auto w : set[u]){
			set[w].erase(u);
			if (set[w].count(v)){
				q.push({w, v});
			}
			else{
				set[w].insert(v);
				set[v].insert(w);
			}
		}
		set[u].clear();
	}
	return printf(sz[find(1)] == n ? "YES\n" : "NO\n"), 0;
}