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>

#define MAXN (100010)

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

int getfather(int x) { return x == fa[x] ? x : fa[x] = getfather(fa[x]); }

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