#include<bits/stdc++.h>
#define mk make_pair
#define N 100005
using namespace std;
int n,fa[N];
int find(int x)
{
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
map<pair<int,int>,int>mp;
queue<pair<int,int> >q;
set<int>s[N];
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 t1,t2;
        scanf("%d%d",&t1,&t2);
        s[t1].insert(t2);s[t2].insert(t1);
        if(t1<t2)swap(t1,t2);
        mp[mk(t1,t2)]++;
        if(mp[mk(t1,t2)]>1)q.push(mk(t1,t2));
    }
    for(int j=1;j<n;j++)
    {
        if(q.empty())return puts("NO"),0;
        int x=find(q.front().first),y=find(q.front().second);q.pop();
        mp[mk(x,y)]=-1000;
        if(s[x].size()>s[y].size())swap(x,y);fa[x]=y;
        for(auto i=s[x].begin();i!=s[x].end();++i)
        {
            int v=(*i);if(v==y)continue;
            s[v].erase(s[v].find(x));s[v].insert(y);s[y].insert(v);
            int t1=v,t2=y;if(t1<t2)swap(t1,t2);
            mp[mk(t1,t2)]++;if(mp[mk(t1,t2)]>1)q.push(mk(t1,t2));
        }
        s[y].erase(x);s[x].clear();
    }
    puts("YES");
    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 mk make_pair
#define N 100005
using namespace std;
int n,fa[N];
int find(int x)
{
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
map<pair<int,int>,int>mp;
queue<pair<int,int> >q;
set<int>s[N];
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 t1,t2;
        scanf("%d%d",&t1,&t2);
        s[t1].insert(t2);s[t2].insert(t1);
        if(t1<t2)swap(t1,t2);
        mp[mk(t1,t2)]++;
        if(mp[mk(t1,t2)]>1)q.push(mk(t1,t2));
    }
    for(int j=1;j<n;j++)
    {
        if(q.empty())return puts("NO"),0;
        int x=find(q.front().first),y=find(q.front().second);q.pop();
        mp[mk(x,y)]=-1000;
        if(s[x].size()>s[y].size())swap(x,y);fa[x]=y;
        for(auto i=s[x].begin();i!=s[x].end();++i)
        {
            int v=(*i);if(v==y)continue;
            s[v].erase(s[v].find(x));s[v].insert(y);s[y].insert(v);
            int t1=v,t2=y;if(t1<t2)swap(t1,t2);
            mp[mk(t1,t2)]++;if(mp[mk(t1,t2)]>1)q.push(mk(t1,t2));
        }
        s[y].erase(x);s[x].clear();
    }
    puts("YES");
    return 0;
}

```