#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 100005 ;
using namespace std ;
int n, d[N << 1] ; 
map <int, map<int, int> > cnt ;
multiset <int> E[N] ;
set <pair<int, int> > Q ;

void add(int x, int y) {
	E[x].insert(y) ;
	int tmp = ++ cnt[x][y] ;
	if (tmp > 1) Q.insert(make_pair(min(x, y), max(x, y))) ;
}

void del(int x, int y) {
	E[x].erase(E[x].find(y)) ;
	int tmp = -- cnt[x][y] ;
	if (tmp < 2) Q.erase(make_pair(min(x, y), max(x, y))) ;
}

bool check() {
	rep(k, 1, n - 1) {
		if (Q.empty()) return false ;
		pair <int, int> u = *Q.begin() ;
		Q.erase(u) ;
		int x = u.first, y = u.second ;
		del(x, y), del(x, y), del(y, x), del(y, x) ;
		if (E[x].size() > E[y].size()) swap(x, y) ;
		std :: multiset <int> :: iterator it ;
		d[0] = 0 ;
		for (it = E[x].begin() ; it != E[x].end() ; it ++) {
			int v = *it ;
			add(y, v), add(v, y) ;
			d[++ d[0]] = v ;
		}
		rep(i, 1, d[0]) {
			del(x, d[i]), del(d[i], x) ;
		}

	}
	return true ;
}

int main() {
	scanf("%d", &n) ;
	int x, y ;
	rep(i, 1, n - 1) {
		scanf("%d%d", &x, &y) ;
		add(x, y), add(y, x) ;
	}
	rep(i, 1, n - 1) {
		scanf("%d%d", &x, &y) ;
		add(x, y), add(y, x) ;
	}
	if (check()) printf("YES\n") ;
		else printf("NO\n") ;
	return 0 ; 
}
### Prompt

Your challenge is to write 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 <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 100005 ;
using namespace std ;
int n, d[N << 1] ; 
map <int, map<int, int> > cnt ;
multiset <int> E[N] ;
set <pair<int, int> > Q ;

void add(int x, int y) {
	E[x].insert(y) ;
	int tmp = ++ cnt[x][y] ;
	if (tmp > 1) Q.insert(make_pair(min(x, y), max(x, y))) ;
}

void del(int x, int y) {
	E[x].erase(E[x].find(y)) ;
	int tmp = -- cnt[x][y] ;
	if (tmp < 2) Q.erase(make_pair(min(x, y), max(x, y))) ;
}

bool check() {
	rep(k, 1, n - 1) {
		if (Q.empty()) return false ;
		pair <int, int> u = *Q.begin() ;
		Q.erase(u) ;
		int x = u.first, y = u.second ;
		del(x, y), del(x, y), del(y, x), del(y, x) ;
		if (E[x].size() > E[y].size()) swap(x, y) ;
		std :: multiset <int> :: iterator it ;
		d[0] = 0 ;
		for (it = E[x].begin() ; it != E[x].end() ; it ++) {
			int v = *it ;
			add(y, v), add(v, y) ;
			d[++ d[0]] = v ;
		}
		rep(i, 1, d[0]) {
			del(x, d[i]), del(d[i], x) ;
		}

	}
	return true ;
}

int main() {
	scanf("%d", &n) ;
	int x, y ;
	rep(i, 1, n - 1) {
		scanf("%d%d", &x, &y) ;
		add(x, y), add(y, x) ;
	}
	rep(i, 1, n - 1) {
		scanf("%d%d", &x, &y) ;
		add(x, y), add(y, x) ;
	}
	if (check()) printf("YES\n") ;
		else printf("NO\n") ;
	return 0 ; 
}
```