#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 3;
const int infint = (int)1e9 + 3;
const ll inf = (ll)1e12;
map< pair<int, int>, int> edge;
set< pair<int, pair<int, int> > > deg;
vector<int> G[MAXN];

int n, deleted[MAXN];

void add(int u, int v)
{
	if(u > v)
		swap(u, v);
	G[u].push_back(v);
	G[v].push_back(u);
	deg.erase({-edge[{u, v}], {u, v}});
	edge[{u, v}]++;
	deg.insert({-edge[{u, v}], {u, v}});
}
void in()
{
	for (int i = 0; i < n - 1; i++)
	{
		int u, v;
		cin >> u >> v;
		add(u, v);
	}
	return;
}
void merge(int u, int v)
{
	deg.erase({-edge[{min(u, v), max(u, v)}], {min(u, v), max(u, v)}});
	if(G[u].size() < G[v].size())
		swap(u, v);
	
	for (auto c : G[v])
		if(!deleted[c] && c != u)
			add(u, c), deg.erase({-edge[{min(v, c), max(v, c)}], {min(v, c), max(v, c)}});
	
	cerr << "# " << u << " " << v << endl;
	deleted[v] = 1;
}
int main()
{
	ios::sync_with_stdio(false); 
	cin.tie(0); cout.tie(0);
	cin >> n;
	in(), in();
	int ted = 0;
	while(!deg.empty() && (*deg.begin()).first < -1)
	{
		pair<int, pair<int, int> > cur = *deg.begin();
		merge(cur.second.first, cur.second.second);
		ted++;
	}
	cout << ((ted == n - 1) ? "YES" : "NO");
}

### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 3;
const int infint = (int)1e9 + 3;
const ll inf = (ll)1e12;
map< pair<int, int>, int> edge;
set< pair<int, pair<int, int> > > deg;
vector<int> G[MAXN];

int n, deleted[MAXN];

void add(int u, int v)
{
	if(u > v)
		swap(u, v);
	G[u].push_back(v);
	G[v].push_back(u);
	deg.erase({-edge[{u, v}], {u, v}});
	edge[{u, v}]++;
	deg.insert({-edge[{u, v}], {u, v}});
}
void in()
{
	for (int i = 0; i < n - 1; i++)
	{
		int u, v;
		cin >> u >> v;
		add(u, v);
	}
	return;
}
void merge(int u, int v)
{
	deg.erase({-edge[{min(u, v), max(u, v)}], {min(u, v), max(u, v)}});
	if(G[u].size() < G[v].size())
		swap(u, v);
	
	for (auto c : G[v])
		if(!deleted[c] && c != u)
			add(u, c), deg.erase({-edge[{min(v, c), max(v, c)}], {min(v, c), max(v, c)}});
	
	cerr << "# " << u << " " << v << endl;
	deleted[v] = 1;
}
int main()
{
	ios::sync_with_stdio(false); 
	cin.tie(0); cout.tie(0);
	cin >> n;
	in(), in();
	int ted = 0;
	while(!deg.empty() && (*deg.begin()).first < -1)
	{
		pair<int, pair<int, int> > cur = *deg.begin();
		merge(cur.second.first, cur.second.second);
		ted++;
	}
	cout << ((ted == n - 1) ? "YES" : "NO");
}

```