#include<bits/stdc++.h>
#define mk make_pair
#define fi first
#define nd second
#define pii pair<int,int>
#define sqr(x) ((x)*(x))
using namespace std;
typedef long long ll;
inline ll read() {ll x = 0; char ch = getchar(), w = 1;while(ch < '0' || ch > '9') {if(ch == '-') w = -1;
ch = getchar();}while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0';ch = getchar();}return x * w;}
void write(ll x) {if(x < 0) putchar('-'), x = -x;if(x > 9) write(x / 10);putchar(x % 10 + '0');}
inline void writeln(ll x) {write(x);puts("");}
const int N = 3000000;
int n, fa[N];
set<int>e[N];
map<pii, int>c;
queue<pii>q;
int find(int x){
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main() {
	int tot = 0;
	n = read();
	for(int i = 1; i < n * 2 - 1; ++i){
		int x = read(), y = read();
		e[x].insert(y);
		e[y].insert(x);
		++c[mk(x, y)];++c[mk(y,x)];
		if(c[mk(x,y)] == 2) q.push(mk(x, y)), ++tot;
	}
	for(int i = 1; i <= n; ++i) fa[i] = i;
	while(!q.empty() && tot < n - 1) {
		pii now = q.front();q.pop();
		int x = find(now.fi), y = find(now.nd);
		if(x == y) continue;
		if(e[x].size() > e[y].size()) swap(x, y);
		fa[x] = y;
		for(set<int>::iterator z = e[x].begin(); z != e[x].end(); ++z) {
			if(*z == y) continue;
			c[mk(*z, x)] = c[mk(x, *z)] = 0;
			++c[mk(*z, y)]; ++c[mk(y, *z)];
			e[y].insert(*z);e[*z].insert(y);
			e[*z].erase(x);
			if(c[mk(y, *z)] == 2) q.push(mk(y, *z)), ++tot;
		}
		e[x].clear();e[y].erase(x);
	}
	if(tot < n - 1) {
		puts("NO");
		return 0;
	}
	puts("YES");
	return 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>
#define mk make_pair
#define fi first
#define nd second
#define pii pair<int,int>
#define sqr(x) ((x)*(x))
using namespace std;
typedef long long ll;
inline ll read() {ll x = 0; char ch = getchar(), w = 1;while(ch < '0' || ch > '9') {if(ch == '-') w = -1;
ch = getchar();}while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0';ch = getchar();}return x * w;}
void write(ll x) {if(x < 0) putchar('-'), x = -x;if(x > 9) write(x / 10);putchar(x % 10 + '0');}
inline void writeln(ll x) {write(x);puts("");}
const int N = 3000000;
int n, fa[N];
set<int>e[N];
map<pii, int>c;
queue<pii>q;
int find(int x){
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main() {
	int tot = 0;
	n = read();
	for(int i = 1; i < n * 2 - 1; ++i){
		int x = read(), y = read();
		e[x].insert(y);
		e[y].insert(x);
		++c[mk(x, y)];++c[mk(y,x)];
		if(c[mk(x,y)] == 2) q.push(mk(x, y)), ++tot;
	}
	for(int i = 1; i <= n; ++i) fa[i] = i;
	while(!q.empty() && tot < n - 1) {
		pii now = q.front();q.pop();
		int x = find(now.fi), y = find(now.nd);
		if(x == y) continue;
		if(e[x].size() > e[y].size()) swap(x, y);
		fa[x] = y;
		for(set<int>::iterator z = e[x].begin(); z != e[x].end(); ++z) {
			if(*z == y) continue;
			c[mk(*z, x)] = c[mk(x, *z)] = 0;
			++c[mk(*z, y)]; ++c[mk(y, *z)];
			e[y].insert(*z);e[*z].insert(y);
			e[*z].erase(x);
			if(c[mk(y, *z)] == 2) q.push(mk(y, *z)), ++tot;
		}
		e[x].clear();e[y].erase(x);
	}
	if(tot < n - 1) {
		puts("NO");
		return 0;
	}
	puts("YES");
	return 0;
}

```