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 ll long long
#define mk make_pair
using namespace std;
 
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9;
  
int n, m, no[N];
set<int> G1[N], G2[N];
queue< pair<int, int> > q;
map< pair<int, int>, int> cnt;

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int x, y;
		scanf("%d %d", &x, &y);
		if (x > y) swap(x, y);
		G1[x].insert(y);
		G1[y].insert(x);
		cnt[mk(x, y)]++;
	}
	for (int i = 1; i < n; i++) {
		int x, y;
		scanf("%d %d", &x, &y);
		if (x > y) swap(x, y);
		G2[x].insert(y);
		G2[y].insert(x);
		cnt[mk(x, y)]++;
		if (cnt[mk(x, y)] == 2) q.push(mk(x, y));
	}
	while (q.size()) {// cout << "/";
		int x = q.front().first;
		int y = q.front().second;
		q.pop(); // cout << x << "," << y << endl;
		if (no[x] || no[y]) continue;
		m++;
		no[y] = 1;
		G1[x].erase(G1[x].find(y));
		G1[y].erase(G1[y].find(x));
		G2[x].erase(G2[x].find(y));
		G2[y].erase(G2[y].find(x));
		for (auto i: G1[y]) {
			int xx = i, yy = x;
			if (xx > yy) swap(xx, yy);
			G1[x].insert(i);
			G1[i].insert(x);
			cnt[mk(xx, yy)]++;
			if (cnt[mk(xx, yy)] == 2) q.push(mk(xx, yy));
		}
		for (auto i: G2[y]) {
			int xx = i, yy = x;
			if (xx > yy) swap(xx, yy);
			G2[x].insert(i);
			G2[i].insert(x);
			cnt[mk(xx, yy)]++;
			if (cnt[mk(xx, yy)] == 2) q.push(mk(xx, yy));
		}
	} 
	if (m == n - 1) puts("YES");
	else puts("NO");
}