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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<queue>
#define pi pair<int,int>
#define mk make_pair
using namespace std;
int num,n,sum,d[101001];
char ch,S[3][101]={"NO","YES"};
multiset<int> s[100010];
multiset<int>::iterator it,it2;
queue<pi> q;
int read()
{
	while(!isdigit(ch=getchar()));
	for(num=ch-'0';isdigit(ch=getchar());num=(num<<1)+(num<<3)+ch-'0');
	return num;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n*2-1;i++)
	{
		int x=read(),y=read();
		it=s[x].lower_bound(y);
		if(it!=s[x].end()&&*it==y)
			q.push(mk(x,y));
		s[x].insert(y);s[y].insert(x);
	}
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;q.pop();
		if(d[x]||d[y])  continue;
		sum++;
		if(s[x].size()>s[y].size()) swap(x,y);
		s[x].erase(y);
		for(it=s[x].begin();it!=s[x].end();it++)
		{
			int k=*it;
			it2=s[y].lower_bound(k);
			if(it2!=s[y].end()&&*it2==k)
				q.push(mk(k,y));
			s[k].insert(y);s[y].insert(k);
		}
		d[x]=1;s[x].clear();
	}
	printf("%s\n",S[(sum==n-1)]);	
	return 0;
}