#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 10;
int n;
queue<pair<int, int> > q;
int fa[N];
int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }
set<int> g[N];

pair<int, int> gid(int x, int y) {
	if(x > y) swap(x, y);
	return make_pair(x, y);
}
map<pair<int, int>, int> cnt;

void lk(int u, int v) {
	g[u].emplace(v);
	g[v].emplace(u);
	int cur = ++ cnt[gid(u, v)];
	if(cur == 2) {
		q.emplace(u, v);
	} else if(cur == 3) {
		puts("NO");
		exit(0);
	}
}

int main() {
	scanf("%d", &n);
	for(int i = 1, u, v ; i <= 2 * (n - 1) ; ++ i) {
		scanf("%d%d", &u, &v);
		lk(u, v);
	}
	for(int i = 1 ; i <= n ; ++ i) {
		fa[i] = i;
	}
	for(int i = 1 ; i < n ; ++ i) {
		if(q.empty()) return puts("NO"), 0;
		int x = get(q.front().first), y = get(q.front().second);
		q.pop();
		if(g[x].size() > g[y].size()) {
			swap(x, y);
		}
		fa[x] = y;
		cnt.erase(gid(x, y));
		g[y].erase(x);
		g[x].erase(y);
		for(auto z: g[x]) {
			z = get(z);
			if(z == y) continue;
			cnt.erase(gid(x, z));
			lk(z, y);
			g[z].erase(x);
			g[x].erase(z);
		}
	}
	puts("YES");
}

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 10;
int n;
queue<pair<int, int> > q;
int fa[N];
int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }
set<int> g[N];

pair<int, int> gid(int x, int y) {
	if(x > y) swap(x, y);
	return make_pair(x, y);
}
map<pair<int, int>, int> cnt;

void lk(int u, int v) {
	g[u].emplace(v);
	g[v].emplace(u);
	int cur = ++ cnt[gid(u, v)];
	if(cur == 2) {
		q.emplace(u, v);
	} else if(cur == 3) {
		puts("NO");
		exit(0);
	}
}

int main() {
	scanf("%d", &n);
	for(int i = 1, u, v ; i <= 2 * (n - 1) ; ++ i) {
		scanf("%d%d", &u, &v);
		lk(u, v);
	}
	for(int i = 1 ; i <= n ; ++ i) {
		fa[i] = i;
	}
	for(int i = 1 ; i < n ; ++ i) {
		if(q.empty()) return puts("NO"), 0;
		int x = get(q.front().first), y = get(q.front().second);
		q.pop();
		if(g[x].size() > g[y].size()) {
			swap(x, y);
		}
		fa[x] = y;
		cnt.erase(gid(x, y));
		g[y].erase(x);
		g[x].erase(y);
		for(auto z: g[x]) {
			z = get(z);
			if(z == y) continue;
			cnt.erase(gid(x, z));
			lk(z, y);
			g[z].erase(x);
			g[x].erase(z);
		}
	}
	puts("YES");
}

```