
#include<bits/stdc++.h>
using namespace std;

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define F first
#define S second

const int maxn = 1e5 + 12, lg = 20;
int n, par[maxn], cnt;
set <int> ad[maxn];
vector <pair <int, int> > q;
map <pair <int, int>, int> mp;

void add(int a, int b) {
	mp[make_pair(a, b)]++;
	mp[make_pair(b, a)]++;
	ad[a].insert(b);
	ad[b].insert(a);
	if(mp[make_pair(a, b)] == 2) q.push_back(make_pair(a, b));
	return;
}

int getpar(int a) {
	if(par[a] == a) return a;
	par[a] = getpar(par[a]);
	return par[a];
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) par[i] = i;
	for (int i = 0; i < 2 * (n - 1); i++) {
		int u, v;
		cin >> u >> v;
		v--, u--;
		add(u, v);
	}
	while(cnt < q.size()) {
		if(q.size() >= n) return cout << "NO\n", 0;
		int u, v;
		u = q[cnt].F, v = q[cnt].S, cnt++;
		v = getpar(v), u = getpar(u);
		if(u == v) continue;
		if(ad[u].size() < ad[v].size()) swap(u, v);
		par[v] = u;
		ad[u].erase(v);
		while(ad[v].size()) {
			int x = *ad[v].begin();
			ad[v].erase(x);
			ad[x].erase(v);
			if (x == u) continue;
			mp[make_pair(x, v)] = mp[make_pair(v, x)] = 0;
			add(x, u);
		}
	}
	if(q.size() == n - 1) return cout << "YES", 0;
	return cout << "NO", 0;
}

### 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;

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define F first
#define S second

const int maxn = 1e5 + 12, lg = 20;
int n, par[maxn], cnt;
set <int> ad[maxn];
vector <pair <int, int> > q;
map <pair <int, int>, int> mp;

void add(int a, int b) {
	mp[make_pair(a, b)]++;
	mp[make_pair(b, a)]++;
	ad[a].insert(b);
	ad[b].insert(a);
	if(mp[make_pair(a, b)] == 2) q.push_back(make_pair(a, b));
	return;
}

int getpar(int a) {
	if(par[a] == a) return a;
	par[a] = getpar(par[a]);
	return par[a];
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) par[i] = i;
	for (int i = 0; i < 2 * (n - 1); i++) {
		int u, v;
		cin >> u >> v;
		v--, u--;
		add(u, v);
	}
	while(cnt < q.size()) {
		if(q.size() >= n) return cout << "NO\n", 0;
		int u, v;
		u = q[cnt].F, v = q[cnt].S, cnt++;
		v = getpar(v), u = getpar(u);
		if(u == v) continue;
		if(ad[u].size() < ad[v].size()) swap(u, v);
		par[v] = u;
		ad[u].erase(v);
		while(ad[v].size()) {
			int x = *ad[v].begin();
			ad[v].erase(x);
			ad[x].erase(v);
			if (x == u) continue;
			mp[make_pair(x, v)] = mp[make_pair(v, x)] = 0;
			add(x, u);
		}
	}
	if(q.size() == n - 1) return cout << "YES", 0;
	return cout << "NO", 0;
}

```