#include <bits/stdc++.h>
using namespace std;
#define maxn 200000
int n, cnt, fa[maxn];
set <int> E[maxn];

int read() {
	int x = 0, k = 1;
	char c; c = getchar();
	while(c < '0' ||c > '9') { if(c == '-') k = -1; c = getchar(); }
	while(c >= '0' &&c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * k;
}

struct node {
	int u, v;
	node(int a = 0, int b = 0) {
		u = a, v = b;
	}
}; queue <node> q;

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

int main() {
	n = read(); int t = (n - 1) * 2;
	for(int i = 1; i <= t; i ++) {
		int u = read(), v = read();
		if(!E[u].count(v)) E[u].insert(v), E[v].insert(u);
		else q.push(node(u, v));
	}
	for(int i = 1; i <= n; i ++) fa[i] = i;
	while(!q.empty()) {
		node now = q.front(); q.pop(); cnt ++;
		int x = find(now.u), y = find(now.v);
		if(E[x].size() < E[y].size()) swap(x, y);
		for(set <int> :: iterator it = E[y].begin(); it != E[y].end(); it ++)
			if(*it != x) {
				int u = x, v = *it;
				E[v].erase(y);
				if(!E[u].count(v)) E[u].insert(v), E[v].insert(u);
				else q.push(node(u, v));
			}
		E[x].erase(y), fa[y] = x;
	}
	if(cnt == n - 1) puts("YES");
	else puts("NO");
	return 0;
}
### Prompt

Please formulate a cpp solution 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;
#define maxn 200000
int n, cnt, fa[maxn];
set <int> E[maxn];

int read() {
	int x = 0, k = 1;
	char c; c = getchar();
	while(c < '0' ||c > '9') { if(c == '-') k = -1; c = getchar(); }
	while(c >= '0' &&c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * k;
}

struct node {
	int u, v;
	node(int a = 0, int b = 0) {
		u = a, v = b;
	}
}; queue <node> q;

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

int main() {
	n = read(); int t = (n - 1) * 2;
	for(int i = 1; i <= t; i ++) {
		int u = read(), v = read();
		if(!E[u].count(v)) E[u].insert(v), E[v].insert(u);
		else q.push(node(u, v));
	}
	for(int i = 1; i <= n; i ++) fa[i] = i;
	while(!q.empty()) {
		node now = q.front(); q.pop(); cnt ++;
		int x = find(now.u), y = find(now.v);
		if(E[x].size() < E[y].size()) swap(x, y);
		for(set <int> :: iterator it = E[y].begin(); it != E[y].end(); it ++)
			if(*it != x) {
				int u = x, v = *it;
				E[v].erase(y);
				if(!E[u].count(v)) E[u].insert(v), E[v].insert(u);
				else q.push(node(u, v));
			}
		E[x].erase(y), fa[y] = x;
	}
	if(cnt == n - 1) puts("YES");
	else puts("NO");
	return 0;
}
```