#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define pii pair<int, int>
#define fi first
#define se second
using namespace std;

const int N = 1e5 + 10;
int n, fa[N], tot;
map <pii, int> mp;
multiset <int> s[N];
queue <pii> q;

pii mk(int x, int y) {
	if (x > y) return pii(x, y);
	return pii(y, x);
}

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

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++ i) fa[i] = i;
	for (int i = 1; i <= n * 2 - 2; ++ i) {
		int x, y;
		scanf("%d %d", &x, &y);
		s[x].insert(y);
		s[y].insert(x);
		mp[mk(x, y)] ++;
	}
	for (map <pii, int> :: iterator it = mp.begin(); it != mp.end(); ++ it)
		if ((*it).se == 2) q.push((*it).fi);
	while (!q.empty()) {
		pii now = q.front(); q.pop();
		int x = gf(now.fi), y = gf(now.se);
		if (x == y) continue;
		tot ++;
		if (s[x].size() < s[y].size()) swap(x, y);
		fa[y] = x;
		for (set <int> :: iterator it = s[y].begin(); it != s[y].end(); ++ it) {
			int u = gf(*it);
			if (u == x) continue;
			s[x].insert(u), s[u].insert(x);
			mp[mk(y, u)] --;
			mp[mk(x, u)] ++;
			if (mp[mk(x, u)] == 2) q.push(mk(x, u));
		}
		s[y].clear();
	}
	puts(tot == n - 1 ? "YES" : "NO");
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define pii pair<int, int>
#define fi first
#define se second
using namespace std;

const int N = 1e5 + 10;
int n, fa[N], tot;
map <pii, int> mp;
multiset <int> s[N];
queue <pii> q;

pii mk(int x, int y) {
	if (x > y) return pii(x, y);
	return pii(y, x);
}

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

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++ i) fa[i] = i;
	for (int i = 1; i <= n * 2 - 2; ++ i) {
		int x, y;
		scanf("%d %d", &x, &y);
		s[x].insert(y);
		s[y].insert(x);
		mp[mk(x, y)] ++;
	}
	for (map <pii, int> :: iterator it = mp.begin(); it != mp.end(); ++ it)
		if ((*it).se == 2) q.push((*it).fi);
	while (!q.empty()) {
		pii now = q.front(); q.pop();
		int x = gf(now.fi), y = gf(now.se);
		if (x == y) continue;
		tot ++;
		if (s[x].size() < s[y].size()) swap(x, y);
		fa[y] = x;
		for (set <int> :: iterator it = s[y].begin(); it != s[y].end(); ++ it) {
			int u = gf(*it);
			if (u == x) continue;
			s[x].insert(u), s[u].insert(x);
			mp[mk(y, u)] --;
			mp[mk(x, u)] ++;
			if (mp[mk(x, u)] == 2) q.push(mk(x, u));
		}
		s[y].clear();
	}
	puts(tot == n - 1 ? "YES" : "NO");
	return 0;
}

```