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> 
#define mp(x,y) make_pair(x,y)
#define pii pair<int,int>
using namespace std;
inline int read(){
	int x=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
const int N=1e5+5;
set<int>e[N];
set<int>::iterator it;
map<pii,int>H;
queue<pii>q;
int n,fa[N];
inline int find(int x){return fa[x]==x ? x:fa[x]=find(fa[x]);}
int main(){
    n=read();
    for(int i=1;i<=n*2-2;i++){
        int x=read(),y=read();
        e[x].insert(y);e[y].insert(x);
        H[mp(x,y)]++;H[mp(y,x)]++;
        if(H[mp(x,y)]==2)q.push(mp(x,y));
    }
    for(int i=1;i<=n;i++)fa[i]=i;
    while(!q.empty()){
    	int x=q.front().first,y=q.front().second;q.pop();
		x=find(x);y=find(y);
    	if(e[x].size()<e[y].size())swap(x,y);
    	fa[y]=x;e[x].erase(y);e[y].erase(x);
    	for(it=e[y].begin();it!=e[y].end();it++){
    		int v=*it;
    		H[mp(y,v)]=H[mp(v,y)]=0;
			e[v].erase(y);
			e[x].insert(v);e[v].insert(x);
			H[mp(x,v)]++;H[mp(v,x)]++;
			if(H[mp(x,v)]==2)q.push(mp(x,v));
		}
		e[y].clear();
	}
	int s=0;
	for(int i=1;i<=n;i++)if(i==fa[i])s++;
	if(s==1)puts("YES");else puts("NO");
    return 0;
}