#include<cstdio>
#include<set>
#include<queue>
#include<map>
using namespace std;

typedef long long ll;
const int MAXN=100005;

int n;
int fa[MAXN];
set<int> s[MAXN];
queue<pair<int,int> > q;
map<ll,int> g;

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

 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(make_pair(u,v));
}

void re(int u,int v){
    s[u].erase(v);
    s[v].erase(u);
    g.erase(getID(u,v));
}

bool solve(){
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<n;i++){
        int u,v;
        if(q.empty()) return 0;
        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);
        fa[u]=v;
        re(u,v);
        for(set<int>::iterator i=s[u].begin(),j;i!=s[u].end();i=j){
            j=i;
            j++;
            int x=*i;
            x=find(x);
            re(u,x);
            addEdge(v,x);
        }
    }
    return 1;
}

int main(){
    scanf("%d",&n);
    int u,v;
    for(int i=1;i<=(n-1)<<1;i++){
        scanf("%d%d",&u,&v);
        addEdge(u,v);
    }
    puts(solve()?"YES":"NO");
}

### 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<set>
#include<queue>
#include<map>
using namespace std;

typedef long long ll;
const int MAXN=100005;

int n;
int fa[MAXN];
set<int> s[MAXN];
queue<pair<int,int> > q;
map<ll,int> g;

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

 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(make_pair(u,v));
}

void re(int u,int v){
    s[u].erase(v);
    s[v].erase(u);
    g.erase(getID(u,v));
}

bool solve(){
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<n;i++){
        int u,v;
        if(q.empty()) return 0;
        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);
        fa[u]=v;
        re(u,v);
        for(set<int>::iterator i=s[u].begin(),j;i!=s[u].end();i=j){
            j=i;
            j++;
            int x=*i;
            x=find(x);
            re(u,x);
            addEdge(v,x);
        }
    }
    return 1;
}

int main(){
    scanf("%d",&n);
    int u,v;
    for(int i=1;i<=(n-1)<<1;i++){
        scanf("%d%d",&u,&v);
        addEdge(u,v);
    }
    puts(solve()?"YES":"NO");
}

```