#include<bits/stdc++.h>
#define N 100009
#define P pair<int,int>
#define mm make_pair
using namespace std;
typedef long long ll;
int n,f[N];
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
int find(int x){return f[x]=f[x]==x?x:find(f[x]);}
set<int>s[N];
map<P,int>mp;
queue<P>q;
inline P cl(int x,int y){return mm(min(x,y),max(x,y));}
inline void add(int u,int v){
    s[u].insert(v);s[v].insert(u);
    P now=cl(u,v);
    if(mp.find(now)==mp.end())mp[now]=1;
    else q.push(now);
}
int main(){
    n=rd();
    int x,y;
    for(int i=1;i<=n*2-2;++i){
        x=rd();y=rd();
        add(x,y);
    }
    int ans=0;
    for(int i=1;i<=n;++i)f[i]=i;
    while(!q.empty()){
        P now=q.front();q.pop(); 
        int u=find(now.first),v=find(now.second);
        if(u==v)continue;
        ans++;
        if(s[u].size()>s[v].size())swap(u,v);
        f[u]=v;
        mp.erase(mm(u,v));s[v].erase(u);
        for(set<int>::iterator it=s[u].begin();it!=s[u].end();++it){
            int to=*it;
            int x=v,y=find(to);
            if(v==y)continue;
            mp.erase(cl(u,y));
            P now=cl(x,y);
            if(mp.find(now)==mp.end())mp[now]=1;
            else q.push(now);
            s[v].insert(y);
        }
        s[u].clear();
    }
    if(ans==n-1)puts("YES");
    else puts("NO");
    return 0;
}
### Prompt

Your challenge is to write 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 N 100009
#define P pair<int,int>
#define mm make_pair
using namespace std;
typedef long long ll;
int n,f[N];
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
int find(int x){return f[x]=f[x]==x?x:find(f[x]);}
set<int>s[N];
map<P,int>mp;
queue<P>q;
inline P cl(int x,int y){return mm(min(x,y),max(x,y));}
inline void add(int u,int v){
    s[u].insert(v);s[v].insert(u);
    P now=cl(u,v);
    if(mp.find(now)==mp.end())mp[now]=1;
    else q.push(now);
}
int main(){
    n=rd();
    int x,y;
    for(int i=1;i<=n*2-2;++i){
        x=rd();y=rd();
        add(x,y);
    }
    int ans=0;
    for(int i=1;i<=n;++i)f[i]=i;
    while(!q.empty()){
        P now=q.front();q.pop(); 
        int u=find(now.first),v=find(now.second);
        if(u==v)continue;
        ans++;
        if(s[u].size()>s[v].size())swap(u,v);
        f[u]=v;
        mp.erase(mm(u,v));s[v].erase(u);
        for(set<int>::iterator it=s[u].begin();it!=s[u].end();++it){
            int to=*it;
            int x=v,y=find(to);
            if(v==y)continue;
            mp.erase(cl(u,y));
            P now=cl(x,y);
            if(mp.find(now)==mp.end())mp[now]=1;
            else q.push(now);
            s[v].insert(y);
        }
        s[u].clear();
    }
    if(ans==n-1)puts("YES");
    else puts("NO");
    return 0;
}
```