#include <iostream>
#include <cstdio>
#include <set>
#include <algorithm>
#define rep(i,a,b) for(int i = a; i <= b; i++)
#define dep(i,a,b) for(int i = a; i >= b; i--)
#define Rep(i,a) for(int i = 0; i < a; i ++)
using namespace std;
const int N = 1e5 + 10;
typedef pair<int, int> pii;
#define mp(a,b) make_pair(a,b)
multiset<pii> to[N];
typedef multiset<pii>::iterator sit;

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

void add_edge(int a, int b, int c);

pii q[N]; int l = 0, r = 0;

void merge(int a, int b) {
	a = find(a), b = find(b);
	if (a == b) return;
	if (to[a].size() < to[b].size()) swap(a, b);
	f[b] = a;
	for(sit it = to[b].begin(); it != to[b].end(); it++) {
		int v = it->first, c = it->second;
		sit tmp = to[v].find(mp(b, c));
		to[v].erase(tmp);
		add_edge(v, a, c);
	}
}

void add_edge(int a, int b, int c) {
	sit it = to[a].find(mp(b, !c));
	if (it == to[a].end()) 
		to[a].insert(mp(b, c)), to[b].insert(mp(a, c));
	else {
		to[a].erase(it);
		it = to[b].find(mp(a, !c)); to[b].erase(it);
		q[r++] = mp(a, b);
	}
}

int main() {
	int n; scanf("%d",&n);
	rep(i,1,n) f[i] = i;
	rep(i,1,n - 1) {
		int a, b;
		scanf("%d%d",&a,&b);
		a = find(a), b = find(b);
		add_edge(a, b, 0);
	}
	rep(i,1,n - 1) {
		int a, b;
		scanf("%d%d",&a,&b);
		a = find(a), b = find(b);
		add_edge(a, b, 1);
	}
	while (l < r) merge(q[l].first, q[l].second), l++;
	rep(i,1,n) if (find(i) != find(1)) {
		printf("NO\n"); return 0;
	}
	printf("YES\n");
	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 <iostream>
#include <cstdio>
#include <set>
#include <algorithm>
#define rep(i,a,b) for(int i = a; i <= b; i++)
#define dep(i,a,b) for(int i = a; i >= b; i--)
#define Rep(i,a) for(int i = 0; i < a; i ++)
using namespace std;
const int N = 1e5 + 10;
typedef pair<int, int> pii;
#define mp(a,b) make_pair(a,b)
multiset<pii> to[N];
typedef multiset<pii>::iterator sit;

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

void add_edge(int a, int b, int c);

pii q[N]; int l = 0, r = 0;

void merge(int a, int b) {
	a = find(a), b = find(b);
	if (a == b) return;
	if (to[a].size() < to[b].size()) swap(a, b);
	f[b] = a;
	for(sit it = to[b].begin(); it != to[b].end(); it++) {
		int v = it->first, c = it->second;
		sit tmp = to[v].find(mp(b, c));
		to[v].erase(tmp);
		add_edge(v, a, c);
	}
}

void add_edge(int a, int b, int c) {
	sit it = to[a].find(mp(b, !c));
	if (it == to[a].end()) 
		to[a].insert(mp(b, c)), to[b].insert(mp(a, c));
	else {
		to[a].erase(it);
		it = to[b].find(mp(a, !c)); to[b].erase(it);
		q[r++] = mp(a, b);
	}
}

int main() {
	int n; scanf("%d",&n);
	rep(i,1,n) f[i] = i;
	rep(i,1,n - 1) {
		int a, b;
		scanf("%d%d",&a,&b);
		a = find(a), b = find(b);
		add_edge(a, b, 0);
	}
	rep(i,1,n - 1) {
		int a, b;
		scanf("%d%d",&a,&b);
		a = find(a), b = find(b);
		add_edge(a, b, 1);
	}
	while (l < r) merge(q[l].first, q[l].second), l++;
	rep(i,1,n) if (find(i) != find(1)) {
		printf("NO\n"); return 0;
	}
	printf("YES\n");
	return 0;
}
```