#include <bits/stdc++.h>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define pii pair<int, int>
#define MP make_pair
#define fi first
#define se second
using namespace std;
typedef long long LL;
const int MAXN = 100005;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

void out(bool ok) {
	puts(ok? "YES" : "NO");
	exit(0);
}

int n;
multiset<int> e[MAXN];
queue<pii> q;

inline void AddEdge(int u, int v) {
	if (u == v) return;
	if (u > v) swap(u, v);
	if (e[u].find(v) != e[u].end()) {
		q.push(MP(u, v));
	}
	e[u].insert(v);
	e[v].insert(u);
}

void init() {
	read(n);
	for (int i = 1, m = n * 2 - 2, u, v; i <= m; ++i) {
		read(u); read(v);
		AddEdge(u, v);
	}
}

void solve() {
	for (int i = 1; i < n; ++i) {
		while (871) {
			if (q.empty()) out(0);
			int u = q.front().fi, v = q.front().se;
			q.pop();
			if (e[u].find(v) == e[u].end()) continue;
			if (e[u].size() < e[v].size()) swap(u, v);
			for (auto p : e[v]) {
				e[p].erase(e[p].find(v));
				AddEdge(u, p);
			}
			e[v].clear();
			break;
		}
	}
	out(1);
}

int main() {
	init();
	solve();
	return 0;
}

### Prompt

Generate 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>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define pii pair<int, int>
#define MP make_pair
#define fi first
#define se second
using namespace std;
typedef long long LL;
const int MAXN = 100005;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

void out(bool ok) {
	puts(ok? "YES" : "NO");
	exit(0);
}

int n;
multiset<int> e[MAXN];
queue<pii> q;

inline void AddEdge(int u, int v) {
	if (u == v) return;
	if (u > v) swap(u, v);
	if (e[u].find(v) != e[u].end()) {
		q.push(MP(u, v));
	}
	e[u].insert(v);
	e[v].insert(u);
}

void init() {
	read(n);
	for (int i = 1, m = n * 2 - 2, u, v; i <= m; ++i) {
		read(u); read(v);
		AddEdge(u, v);
	}
}

void solve() {
	for (int i = 1; i < n; ++i) {
		while (871) {
			if (q.empty()) out(0);
			int u = q.front().fi, v = q.front().se;
			q.pop();
			if (e[u].find(v) == e[u].end()) continue;
			if (e[u].size() < e[v].size()) swap(u, v);
			for (auto p : e[v]) {
				e[p].erase(e[p].find(v));
				AddEdge(u, p);
			}
			e[v].clear();
			break;
		}
	}
	out(1);
}

int main() {
	init();
	solve();
	return 0;
}

```