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>
using namespace std;

const int N = 100005;

int n;
multiset <int> conn[N];
map <int,int> e[N];

void erase(int u, int v) {
	multiset<int>::iterator it = conn[u].find(v);
	conn[u].erase(it);
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i <= 2 * n - 2; ++i) {
		int u, v; cin >> u >> v;
		e[u][v]++; e[v][u]++;
		conn[u].insert(v); conn[v].insert(u);
	}
	queue < pair<int,int> > q;
	for (int i = 1; i <= n; ++i) {
		for (map<int,int>::iterator it = e[i].begin(); it != e[i].end(); ++it) {
			if (it->first > i && e[i][it->first] >= 2) q.push(make_pair(i, it->first));
		}
 	}

 	while(!q.empty()) {
 		int u = q.front().first, v = q.front().second; q.pop();
 		if (conn[u].empty() || conn[v].empty()) continue;
 		if (e[u][v] > 2) return printf("NO\n"), 0;
 		if (conn[u].size() < conn[v].size()) swap(u, v);
 		erase(u, v); erase(u, v); e[u][v] -= 2;
 		erase(v, u); erase(v, u); e[v][u] -= 2;

 		for (auto w : conn[v]) {
 			erase(w, v); e[w][v]--; e[v][w]--;
 			conn[w].insert(u); e[w][u]++;
 			conn[u].insert(w); e[u][w]++;
 			if (e[u][w] >= 2) q.push(make_pair(u,w));
 		}
 		conn[v].clear();
 	}

 	for (int i = 1; i <= n; ++i) if (!conn[i].empty()) return printf("NO\n"), 0;
 	printf("YES\n");
}