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 <bits/stdc++.h>
using namespace std;
const int N=1e5+7;
struct edge {
	int t,nxt;
}e[N];
int n,num;
int fa[N];
#define pii pair<int,int>
#define mk make_pair
queue<pii>q;
map<pii,int>mp;
set<int>s[N];
int find(int x) {return x==fa[x]?x:fa[x]=find(fa[x]);}
int main() {
	scanf("%d",&n);num=n;
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1,u,v;i<2*n-1;i++) {
		scanf("%d%d",&u,&v);
		s[u].insert(v),s[v].insert(u);
		mp[mk(u,v)]++,mp[mk(v,u)]++;
		if (mp[mk(u,v)]==2) q.push(mk(u,v));
	}
	while (!q.empty()) {
		pii now=q.front();q.pop();
		int u=find(now.first),v=find(now.second);
		num--;fa[v]=u;
		s[u].erase(v);
		set<int>::iterator it;
		for (it=s[v].begin();it!=s[v].end();it++) {
			int to=*it;
			if (to==u) continue;
			s[to].erase(v);
			s[u].insert(to);s[to].insert(u);
			mp[mk(u,to)]++;mp[mk(to,u)]++;
			if (mp[mk(u,to)]==2) q.push(mk(u,to));
		}
		s[v].clear();
	}
	if (num>1) puts("NO");
	else puts("YES");
}