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 <iostream>
#include <cstdio>
#include <set>
#include <queue>
#include <unordered_map>

#define X first
#define Y second
#define mp std::make_pair
#define MAX_N 100010
typedef std::pair <int, int> pii;

int n, sz[MAX_N], fa[MAX_N];
std::set <int> S[MAX_N];
std::unordered_map <int, std::unordered_map<int, int> > cnt;
std::queue <pii> q;

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

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		sz[i] = 1, fa[i] = i;
	for(int i = 1, x, y; i <= 2 * n - 2; i++){
		scanf("%d%d", &x, &y);
		S[x].insert(y); S[y].insert(x);
		cnt[x][y]++, cnt[y][x]++;
		if(cnt[x][y] == 2) q.push(mp(x, y));
	}
	while(!q.empty()){
		int x = q.front().X, y = q.front().Y;
		q.pop();
		x = getfa(x), y = getfa(y);
		if(x == y) continue;
		S[x].erase(y), S[y].erase(x);
		if(sz[x] < sz[y]) std::swap(x, y);
		for(auto i: S[y]){
			S[i].erase(y); S[i].insert(x);
			S[x].insert(i);
			cnt[i][x]++, cnt[x][i]++;
			cnt[i][y] = cnt[y][i] = 0;
			if(cnt[i][x] == 2) 
				q.push(mp(i, x));
		}
		sz[x] += sz[y], fa[y] = x; 
		S[y].clear();
	}
	int cnt = 0;
	for(int i = 1; i <= n; i++) 
		cnt += fa[i] == i;
	puts(cnt > 1 ? "NO" : "YES");
	return 0;
}