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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define maxn 400010
using namespace std;
template<typename T> inline void read(T &x)
{
    x=0;char c=getchar();bool flag=false;
    while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    if(flag)x=-x;
}
int n,cnt;
int fa[maxn];
set<int> e[maxn];
queue<pair<int,int> > q;
int find(int x)
{
    return fa[x]==x?x:fa[x]=find(fa[x]);
}
void add(int x,int y)
{
    if(!e[x].count(y)) e[x].insert(y),e[y].insert(x);
    else q.push(make_pair(x,y)),cnt++;
}
int main()
{
    read(n);
    for(int i=1;i<=n;++i) fa[i]=i;
    for(int i=1;i<=2*n-2;++i)
    {
        int x,y;
        read(x),read(y),add(x,y);
    }
    while(!q.empty())
    {
        int x=find(q.front().first),y=find(q.front().second);
        q.pop();
        if(e[x].size()>e[y].size()) swap(x,y);
        fa[x]=y,e[y].erase(x);
        for(set<int>::iterator it=e[x].begin();it!=e[x].end();++it)
        {
            int p=*it;
            if(p==y) continue;
            e[p].erase(x),add(p,y);
        }
    }
    puts(cnt==n-1?"YES":"NO");
    return 0;
}