#include <cstdio>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define N 100010

using namespace std;

typedef pair<int,int> par;

int n;
int fa[N];
map<par,int> M,A;
multiset<int> S[N];
queue<par> Q;

inline par makep(int x,int y){
    if(x>y) return par(y,x);
    return par(x,y);
}

int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<=2*n-2;i++){
        int x,y; par z;
        scanf("%d%d",&x,&y);
        z=makep(x,y);
        S[x].insert(y); S[y].insert(x);
        if((++M[z])>1&&!A.count(z))
            A[z]=1,Q.push(z);
    }
    for(int k=1;k<n;k++){
        int x,y;
        while(1){
            if(Q.empty()) return puts("NO"),0;
            x=find(Q.front().first),y=find(Q.front().second); Q.pop();
            if(x!=y) break;
        }
        if(S[x].size()>S[y].size()) swap(x,y); fa[x]=y;
        for(set<int>::iterator i=S[x].begin();i!=S[x].end();i++){
            int v=find(*i);
            if(v==y) continue;
            S[y].insert(v); S[v].insert(y);
            if((++M[makep(y,v)])>1&&!A.count(makep(y,v)))
                Q.push(makep(y,v)),A[makep(y,v)]=1;
            S[v].erase(S[v].find(x));
        }
        S[y].erase(x);
        S[x].clear();
    }
    puts("YES");
    return 0;
}
### Prompt

Your task is to create 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#define N 100010

using namespace std;

typedef pair<int,int> par;

int n;
int fa[N];
map<par,int> M,A;
multiset<int> S[N];
queue<par> Q;

inline par makep(int x,int y){
    if(x>y) return par(y,x);
    return par(x,y);
}

int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<=2*n-2;i++){
        int x,y; par z;
        scanf("%d%d",&x,&y);
        z=makep(x,y);
        S[x].insert(y); S[y].insert(x);
        if((++M[z])>1&&!A.count(z))
            A[z]=1,Q.push(z);
    }
    for(int k=1;k<n;k++){
        int x,y;
        while(1){
            if(Q.empty()) return puts("NO"),0;
            x=find(Q.front().first),y=find(Q.front().second); Q.pop();
            if(x!=y) break;
        }
        if(S[x].size()>S[y].size()) swap(x,y); fa[x]=y;
        for(set<int>::iterator i=S[x].begin();i!=S[x].end();i++){
            int v=find(*i);
            if(v==y) continue;
            S[y].insert(v); S[v].insert(y);
            if((++M[makep(y,v)])>1&&!A.count(makep(y,v)))
                Q.push(makep(y,v)),A[makep(y,v)]=1;
            S[v].erase(S[v].find(x));
        }
        S[y].erase(x);
        S[x].clear();
    }
    puts("YES");
    return 0;
}
```