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 <cstdio>
#include <set>

using namespace std;
const int N=100005;
int fa[N],stx[N],sty[N],n,top=0;
set<int> s[N];

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

void insert(int x,int y){
	if (s[x].find(y)!=s[x].end()){
		stx[++top]=x;sty[top]=y;
		s[x].erase(y);
		s[y].erase(x);
		return;
	}
	s[x].insert(y);
	s[y].insert(x);
}

void erase(int x,int y){
	x=find(x);y=find(y);
	if (x==y) return;
	if (s[x].size()<s[y].size()) swap(x,y);
	fa[y]=x;
	for (set<int>::iterator it=s[y].begin();it!=s[y].end();it++){
		int z=find(*it);s[z].erase(y);
		if (z!=x && z!=y) insert(x,z);
	}
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n*2-2;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		insert(x,y);
	}
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1;i<=top;i++) erase(stx[i],sty[i]);
	if (top==n-1) printf("YES\n");
		else printf("NO\n");
}