#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int,int> pi;
const int N=1e5+10;
int n,u,v,cnt,g[N];queue<pi> q;
map<pi,int> a;set<int> f[N];
int gfa(int x){return (g[x]^x)?gfa(g[x]):x;}
int main(){
    scanf("%d",&n);
    rep(i,1,n) g[i]=i;
    rep(i,1,2*n-2){
        scanf("%d%d",&u,&v),f[u].insert(v),f[v].insert(u),++a[mp(u,v)],++a[mp(v,u)];
        if(a[mp(u,v)]==2) q.push(mp(u,v));
    }
    while(!q.empty()){
        int u=q.front().fi,v=q.front().se;q.pop();
        u=gfa(u),v=gfa(v);
        if(f[u].size()<f[v].size()) swap(u,v);
        f[u].erase(v);f[v].erase(u);
        for(int x:f[v]){
            f[x].erase(v);f[u].insert(x);f[x].insert(u);
            ++a[mp(u,x)];++a[mp(x,u)];a[mp(v,x)]=a[mp(x,v)]=0;
            if(a[mp(u,x)]==2) q.push(mp(u,x));
        }++cnt;g[v]=u;
    }
    puts(cnt==n-1?"YES":"NO");
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int,int> pi;
const int N=1e5+10;
int n,u,v,cnt,g[N];queue<pi> q;
map<pi,int> a;set<int> f[N];
int gfa(int x){return (g[x]^x)?gfa(g[x]):x;}
int main(){
    scanf("%d",&n);
    rep(i,1,n) g[i]=i;
    rep(i,1,2*n-2){
        scanf("%d%d",&u,&v),f[u].insert(v),f[v].insert(u),++a[mp(u,v)],++a[mp(v,u)];
        if(a[mp(u,v)]==2) q.push(mp(u,v));
    }
    while(!q.empty()){
        int u=q.front().fi,v=q.front().se;q.pop();
        u=gfa(u),v=gfa(v);
        if(f[u].size()<f[v].size()) swap(u,v);
        f[u].erase(v);f[v].erase(u);
        for(int x:f[v]){
            f[x].erase(v);f[u].insert(x);f[x].insert(u);
            ++a[mp(u,x)];++a[mp(x,u)];a[mp(v,x)]=a[mp(x,v)]=0;
            if(a[mp(u,x)]==2) q.push(mp(u,x));
        }++cnt;g[v]=u;
    }
    puts(cnt==n-1?"YES":"NO");
    return 0;
}
```