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 <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
const int maxn = 100005;
typedef pair<int, int> pii;
int N, f[maxn];
set<int> st[maxn];
queue<pii> q;
int find(int x) {
	return f[x] == x ? x : f[x] = find(f[x]);
}
bool merge(int x, int y) {
	// printf("%d %d\n", x, y);
	x = find(x), y = find(y);
	if(x == y) return false;
	if(st[x].size() < st[y].size()) swap(x, y);
	f[y] = x;
	for(set<int>::iterator it = st[y].begin(); 
		it != st[y].end(); ++it) {
		int t = find(*it);
		if(x == t) continue;
		if(st[x].count(t)) q.push(pii(x, t));
		else st[x].insert(t), st[t].insert(x);
	}
	st[y].clear(), st[x].erase(y);
	return true;
}
int main() {
	scanf("%d", &N);
	for(int i = 1, u, v; i < N + N - 1; ++i) {
		scanf("%d%d", &u, &v);
		if(st[u].count(v)) q.push(pii(u, v));
		st[u].insert(v), st[v].insert(u);
	}
	for(int i = 1; i <= N; ++i) f[i] = i;
	int cnt = 0;
	while(!q.empty()) {
		pii x = q.front(); q.pop();
		cnt += merge(x.first, x.second);
	}
	puts(cnt == N - 1 ? "YES" : "NO");
	getchar(), getchar();
	return 0;
}