#include<bits/stdc++.h>
#define maxn 100009
#define mk make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef pair<int,int> P;
multiset<int>st[maxn];
map<P,int>mp;
map<P,int>::iterator it;
multiset<int>::iterator jt;
int fa[maxn],sz[maxn];
queue<P>q;
vector<int>vec[maxn];
int find(int x){if(fa[x]!=x) fa[x]=find(fa[x]);return fa[x];}
int main(){
  int n,x,y;
  scanf("%d",&n);
  for(int i=1;i<n;i++){
    scanf("%d%d",&x,&y);
    st[x].insert(y),st[y].insert(x);
    mp[mk(x,y)]+=1,mp[mk(y,x)]+=1;
  }
  for(int i=1;i<n;i++){
    scanf("%d%d",&x,&y);
    st[x].insert(y),st[y].insert(x);
    mp[mk(x,y)]+=1,mp[mk(y,x)]+=1;
  }
  for(it=mp.begin();it!=mp.end();it++)
    if(it->se==2) q.push(it->fi);
  for(int i=1;i<=n;i++) fa[i]=i,sz[i]=1,vec[i].pb(i);
  for(int T=1;T<n;T++){
  tp:
    if(q.empty()) puts("NO"),exit(0);
    P u=q.front();q.pop();
    mp.erase(u);
    int x=find(u.fi),y=find(u.se);
    if(x==y) goto tp;
    if(sz[x]>sz[y]) swap(x,y);
    for(jt=st[x].begin();jt!=st[x].end();jt++){
      int k=find(*jt);if(k==y) continue;
      st[y].insert(k),st[k].insert(y),mp[mk(y,k)]+=1,mp[mk(k,y)]+=1;
      if(mp[mk(y,k)]==2) q.push(mk(y,k));
      if(mp[mk(k,y)]==2) q.push(mk(k,y));
    }
    fa[x]=y,sz[y]+=sz[x];
  }
  puts("YES");
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define maxn 100009
#define mk make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef pair<int,int> P;
multiset<int>st[maxn];
map<P,int>mp;
map<P,int>::iterator it;
multiset<int>::iterator jt;
int fa[maxn],sz[maxn];
queue<P>q;
vector<int>vec[maxn];
int find(int x){if(fa[x]!=x) fa[x]=find(fa[x]);return fa[x];}
int main(){
  int n,x,y;
  scanf("%d",&n);
  for(int i=1;i<n;i++){
    scanf("%d%d",&x,&y);
    st[x].insert(y),st[y].insert(x);
    mp[mk(x,y)]+=1,mp[mk(y,x)]+=1;
  }
  for(int i=1;i<n;i++){
    scanf("%d%d",&x,&y);
    st[x].insert(y),st[y].insert(x);
    mp[mk(x,y)]+=1,mp[mk(y,x)]+=1;
  }
  for(it=mp.begin();it!=mp.end();it++)
    if(it->se==2) q.push(it->fi);
  for(int i=1;i<=n;i++) fa[i]=i,sz[i]=1,vec[i].pb(i);
  for(int T=1;T<n;T++){
  tp:
    if(q.empty()) puts("NO"),exit(0);
    P u=q.front();q.pop();
    mp.erase(u);
    int x=find(u.fi),y=find(u.se);
    if(x==y) goto tp;
    if(sz[x]>sz[y]) swap(x,y);
    for(jt=st[x].begin();jt!=st[x].end();jt++){
      int k=find(*jt);if(k==y) continue;
      st[y].insert(k),st[k].insert(y),mp[mk(y,k)]+=1,mp[mk(k,y)]+=1;
      if(mp[mk(y,k)]==2) q.push(mk(y,k));
      if(mp[mk(k,y)]==2) q.push(mk(k,y));
    }
    fa[x]=y,sz[y]+=sz[x];
  }
  puts("YES");
  return 0;
}

```