#include <bits/stdc++.h>
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii; typedef set<int> si; typedef set<int>::iterator sit; const int N=100005;
int n; int bl[N]; si s[N]; queue<pii> q; map<ll,int> g;
inline void swap(int &x,int &y){ x^=y^=x^=y; }   
int find(int u){ return bl[u]==u?u:(bl[u]=find(bl[u])); }
inline ll getID(int x,int y){ if(x>y) swap(x,y); return 1ll*n*y+x; }
void addEdge(int u,int v){ s[u].insert(v); s[v].insert(u); ll eid=getID(u,v); int t=g[eid]+1; g[eid]++; if(t==2) q.push(mp(u,v)); }
void removeEdge(int u,int v){ s[u].erase(v); s[v].erase(u); g.erase(getID(u,v)); }
void readData(){ scanf("%d",&n); int u,v; for(int i=1;i<=(n-1)<<1;i++){ scanf("%d%d",&u,&v); addEdge(u,v); } }
bool solve(){
    for(int i=1;i<=n;i++) bl[i]=i;
    for(int i=1;i<n;i++){
        int u,v; if(q.empty()) return false;
        u=q.front().first; v=q.front().second; q.pop(); u=find(u); v=find(v); if(s[u].size()>s[v].size()) swap(u,v); bl[u]=v; removeEdge(u,v);
        for(sit i=s[u].begin(),j;i!=s[u].end();i=j){ j=i; j++; int x=*i; x=find(x); removeEdge(u,x); addEdge(v,x); }
    }
    return true;
}
int main(){ readData(); puts(solve()?"YES":"NO"); return 0; }
### Prompt

Develop a solution in Cpp to the problem described below:

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 mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii; typedef set<int> si; typedef set<int>::iterator sit; const int N=100005;
int n; int bl[N]; si s[N]; queue<pii> q; map<ll,int> g;
inline void swap(int &x,int &y){ x^=y^=x^=y; }   
int find(int u){ return bl[u]==u?u:(bl[u]=find(bl[u])); }
inline ll getID(int x,int y){ if(x>y) swap(x,y); return 1ll*n*y+x; }
void addEdge(int u,int v){ s[u].insert(v); s[v].insert(u); ll eid=getID(u,v); int t=g[eid]+1; g[eid]++; if(t==2) q.push(mp(u,v)); }
void removeEdge(int u,int v){ s[u].erase(v); s[v].erase(u); g.erase(getID(u,v)); }
void readData(){ scanf("%d",&n); int u,v; for(int i=1;i<=(n-1)<<1;i++){ scanf("%d%d",&u,&v); addEdge(u,v); } }
bool solve(){
    for(int i=1;i<=n;i++) bl[i]=i;
    for(int i=1;i<n;i++){
        int u,v; if(q.empty()) return false;
        u=q.front().first; v=q.front().second; q.pop(); u=find(u); v=find(v); if(s[u].size()>s[v].size()) swap(u,v); bl[u]=v; removeEdge(u,v);
        for(sit i=s[u].begin(),j;i!=s[u].end();i=j){ j=i; j++; int x=*i; x=find(x); removeEdge(u,x); addEdge(v,x); }
    }
    return true;
}
int main(){ readData(); puts(solve()?"YES":"NO"); return 0; }
```