Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1 3
3 2

Sample Output 1

YES

The objective is achievable, as below:

First, select the path connecting vertices 1 and 3, and remove a blue edge 1-2. Then, span a new red edge 1-3.

Next, select the path connecting vertices 2 and 3, and remove a blue edge 2-3. Then, span a new red edge 2-3.

Sample Input 2

5
1 2
2 3
3 4
4 5
3 4
2 4
1 4
1 5

Sample Output 2

YES

Sample Input 3

6
1 2
3 5
4 6
1 6
5 1
5 3
1 4
2 6
4 3
5 6

Sample Output 3

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cmath>
#include<math.h>
#include<ctype.h>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<complex>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cwchar>
#include<cwctype>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
int n,fa[100010];
map<long long,int> mp;
set<int> s[100010];
queue<long long> q;
int find(int x)
{
	if (fa[x]!=x)
	{
		fa[x]=find(fa[x]);
	}
	return fa[x];
}
long long get(int x,int y)
{
	if (x>y)
	{
		swap(x,y);
	}
	return 1LL*x*n+1LL*y;
}
void add(int x,int y)
{
	s[x].insert(y);
	s[y].insert(x);
	long long now=get(x,y);
	mp[now]++;
	if (mp[now]==3)
	{
		puts("NO");
		exit(0);
	}
	if (mp[now]==2)
	{
		q.push(now);
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	for (int i=0;i<2*(n-1);i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);
	}
	for (int i=1;i<n;i++)
	{
		if (q.empty())
		{
			puts("NO");
			return 0;
		}
		long long now=q.front();
		q.pop();
		int x=(now-1)/(long long)n;
		int y=(now-1)%(long long)n+1;
		//cout<<now<<" "<<x<<" "<<y<<endl;
		s[x].erase(y);
		s[y].erase(x);
		mp.erase(get(x,y));
		if (s[x].size()>s[y].size())
		{
			swap(x,y);
		}
		fa[x]=y;
		for (auto v:s[x])
		{
			//cout<<x<<" "<<v<<endl;
			v=find(v);
			if (v==y)
			{
				continue;
			}
			s[v].erase(x);
			s[x].erase(v);
			mp.erase(get(x,v));
			add(v,y);
		}
	}
	puts("YES");
    return 0;
}
//流水淡，碧天长，红尘一刹风乱，吹的散紫薇花瓣，吹不散因缘红线。