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 maxn = 100010;

set<int> s[maxn];
queue<pair<int, int> > q;

int f[maxn];
int find(int x) {
	return f[x] == x ? x : f[x] = find(f[x]);
}

int main() {
	int n; scanf("%d", &n);
	for(int i = 1; i <= n; ++i) f[i] = i;
	for(int i = 0; i < (n - 1) << 1; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		if(s[u].count(v)) {
			q.push({u, v});
		} else {
			s[u].insert(v);
			s[v].insert(u);
		}
	}
	while(!q.empty()) {
		int u, v;
		tie(u, v) = q.front(); q.pop();
		u = find(u), v = find(v);
		if(s[u].size() < s[v].size()) swap(u, v);
		s[u].erase(v);
		s[v].erase(u);
		for(auto x: s[v]) {
			s[x].erase(v);
			if(s[x].count(u)) {
				q.push({u, x});
			} else {
				s[x].insert(u);
				s[u].insert(x);
			}
		}
		s[v].clear();
		f[v] = u;
	}
	int root = find(1);
	bool flag = true;
	for(int i = 2; i <= n; ++i)
		if(find(i) != root) flag = false;
	puts(flag ? "YES" : "NO");
	return 0;
}