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 fi first
#define se second
#define mkp std::make_pair

typedef std::pair <int, int> pii;

const int kN = 1e5 + 5;

int n, fa[kN];
std::queue <pii> que;
std::set <int> E[kN];
std::map <int, int> cnt[kN];
int Find(int x) { return fa[x] == x ? x : fa[x] = Find(fa[x]); }
bool Merge(int u, int v) {
	u = Find(u), v = Find(v);
	if(u == v) return false;
	if(E[u].size() < E[v].size()) std::swap(u, v);
	for(auto i : E[v]) if(i != u) {
		E[u].insert(i); ++cnt[u][i];
		if(cnt[u][i] >= 2) que.push(mkp(u, i));
		E[i].erase(v); --cnt[i][v];
		E[i].insert(u); ++cnt[i][u];
	}
	fa[v] = u;
	return true;
}
bool Solve() {
	int res = n - 1;
	while(!que.empty()) {
		pii cur = que.front(); que.pop();
		res -= Merge(cur.fi, cur.se);
	}
	return !res;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) fa[i] = i;
	for(int i = 1; i < 2 * n - 1; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		E[u].insert(v); E[v].insert(u);
		++cnt[u][v]; ++cnt[v][u];
		if(cnt[u][v] >= 2) {
			que.push(mkp(u, v));
		}
	}
	printf(Solve() ? "YES\n" : "NO\n");
	return 0; 
}