#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#define N 100005
#define Pii pair<int, int>
#define fi first
#define se second
using namespace std;
inline int read(){
	int sum = 0, t = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') t = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ sum = sum * 10 + ch - '0'; ch = getchar(); }
	return sum * t;
}

int n, tot = 1;
map<Pii, int>ret;
vector<int>E[N];
queue<Pii>q;

Pii mk(int x, int y){ if(x > y) swap(x, y); return make_pair(x, y); }

int fa[N], sz[N];

void add(int x, int y){
	E[x].push_back(y); E[y].push_back(x);
	ret[mk(x, y)]++;
}

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

int main(){
	n = read();
	for(int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1;
	for(int i = 1; i <= n * 2 - 2; i++) add(read(), read());
	for(map<Pii, int>::iterator it = ret.begin(); it != ret.end(); it++)
		if(it->se == 2) q.push(it->fi);
	while(!q.empty()){
		Pii u = q.front(); q.pop();
		int x = find(u.fi), y = find(u.se);
		if(x == y) continue ;
		++tot;
		if(sz[x] < sz[y]) swap(x, y);
		for(int i = 0; i < E[y].size(); i++){
			int t = find(E[y][i]);
			if(t == x) continue ;
			E[x].push_back(t);
			ret[mk(y, t)]--;
			ret[mk(x, t)]++;
			if(ret[mk(x, t)] == 2) q.push(mk(x, t));
		}
		E[y].clear();
		sz[x] += sz[y];
		fa[y] = x;
	}
	puts(tot == n ? "YES" : "NO");
}

### Prompt

Develop a solution in Cpp 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<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#define N 100005
#define Pii pair<int, int>
#define fi first
#define se second
using namespace std;
inline int read(){
	int sum = 0, t = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') t = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ sum = sum * 10 + ch - '0'; ch = getchar(); }
	return sum * t;
}

int n, tot = 1;
map<Pii, int>ret;
vector<int>E[N];
queue<Pii>q;

Pii mk(int x, int y){ if(x > y) swap(x, y); return make_pair(x, y); }

int fa[N], sz[N];

void add(int x, int y){
	E[x].push_back(y); E[y].push_back(x);
	ret[mk(x, y)]++;
}

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

int main(){
	n = read();
	for(int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1;
	for(int i = 1; i <= n * 2 - 2; i++) add(read(), read());
	for(map<Pii, int>::iterator it = ret.begin(); it != ret.end(); it++)
		if(it->se == 2) q.push(it->fi);
	while(!q.empty()){
		Pii u = q.front(); q.pop();
		int x = find(u.fi), y = find(u.se);
		if(x == y) continue ;
		++tot;
		if(sz[x] < sz[y]) swap(x, y);
		for(int i = 0; i < E[y].size(); i++){
			int t = find(E[y][i]);
			if(t == x) continue ;
			E[x].push_back(t);
			ret[mk(y, t)]--;
			ret[mk(x, t)]++;
			if(ret[mk(x, t)] == 2) q.push(mk(x, t));
		}
		E[y].clear();
		sz[x] += sz[y];
		fa[y] = x;
	}
	puts(tot == n ? "YES" : "NO");
}

```