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:
//A,Elebereth Gilthoniel mantae!
//For Temeria!
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<vector>
#include<set>
#include<cstdio>
#include<map>
#include<stack>
#include<queue>
#include<cstring> 
using namespace std;
long long n;vector<long long> a,b;map<long long,long long> h;set<long long> f[100010]; 
map<long long,long long> vis;
int main()
{
	cin>>n;
	for(long long i=0;i<n+n-1-1;i++)
	{
		long long x,y;
		cin>>x>>y;h[x*n+y]++,h[y*n+x]++;
		f[x].insert(y),f[y].insert(x);
		if(h[x*n+y]==2) a.push_back(x),b.push_back(y);
	}
	long long kl=0;
	for(long long i=0;i<a.size();i++)
	{
		if(h[a[i]*n+b[i]]==2)
		{++kl;
			if(f[a[i]].size()<f[b[i]].size()) swap(a[i],b[i]);
			while(f[b[i]].size())
			{
				long long t=*f[b[i]].begin();
				h[a[i]*n+t]+=h[b[i]*n+t];
				h[t*n+a[i]]+=h[b[i]*n+t];
				h[b[i]*n+t]=h[t*n+b[i]]=0;
				f[a[i]].insert(t);f[t].erase(b[i]);
				f[t].insert(a[i]);f[b[i]].erase(t);
				if(h[a[i]*n+t]==2)a.push_back(a[i]),b.push_back(t);
			}
		}
	}
	if(kl==n-1) cout<<"YES";else cout<<"NO";return 0;
}