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<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#define mp(x,y) make_pair(x,y)
using namespace std;
typedef pair<int,int> pi;
const int N=100005;
int n,f[N];
set<int>se[N];
map<pi,int>ma;
queue<pi>que;
inline int read(){
	int x=0,f=0,c=getchar();
	for(;c>'9'||c<'0';f=c=='-',c=getchar());
	for(;c>='0'&&c<='9';c=getchar())
	x=(x<<1)+(x<<3)+c-'0';return f?-x:x;
}
inline int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int main(){
    int i,x,y;
    n=read();
    for(i=1;i<=n+n-2;i++){
        x=read();y=read();
        se[x].insert(y);se[y].insert(x);
        ma[mp(x,y)]++;ma[mp(y,x)]++;
        if(ma[mp(x,y)]==2)que.push(mp(x,y));
    }
    for(i=1;i<=n;i++)f[i]=i;
    set<int>::iterator it;
    while(!que.empty()){
        x=que.front().first;y=que.front().second;
		que.pop();x=find(x);y=find(y);
        se[x].erase(y);se[y].erase(x);
        if (se[x].size()<se[y].size())swap(x,y);
        for (it=se[y].begin();it!=se[y].end();it++){
            int to=(*it);
            ma[mp(to,y)]=ma[mp(y,to)]=0;
            se[to].erase(y);se[to].insert(x);
            ma[mp(to,x)]++;ma[mp(x,to)]++;
            if(ma[mp(to,x)]==2)que.push(mp(to,x));
            se[x].insert(to);
        }f[y]=x;se[y].clear();
    }
    int s=0;
    for(i=1;i<=n;i++)s+=f[i]==i;
    puts(s==1?"YES":"NO");
    return 0;
}