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 <set>
#include <map>
#include <queue>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef pair<int,int> pii;

const int maxn=1e5+5;

int n;
int fa[maxn];
queue<pii> q;
set<int>s[maxn];
map<pii,int>cnt;
set<int>::iterator it;

int read() {
	int x=0,f=1;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return x*f;
}

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

int main() {
	n=read();
	for(int i=1;i<2*n-1;i++) {
		int x=read(),y=read();
		s[x].insert(y),s[y].insert(x);
		if(x>y)swap(x,y);
		pii tmp=make_pair(x,y);
		cnt[tmp]++;
		if(cnt[tmp]==2)q.push(tmp);
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<n;i++) {
		if(q.empty()) {puts("NO");return 0;}
		int x=0,y=0;
		for(;x==y;x=find(q.front().first),y=find(q.front().second),q.pop());
		if(s[x].size()>s[y].size())swap(x,y);fa[x]=y;
		for(it=s[x].begin();it!=s[x].end();it++) {
			int v=*s[x].find(*it);
			s[v].erase(s[v].find(x));
			if(v==y)continue;
			s[v].insert(y),s[y].insert(v);
			pii tmp;
			if(v>y)tmp=make_pair(y,v);
			else tmp=make_pair(v,y);
			cnt[tmp]++;
			if(cnt[tmp]==2)q.push(tmp);
		}
	}
	puts("YES");
	return 0;
}