#include <cstdio>
#include <algorithm>
#include <cctype>
#include <set>
#include <map>
#include <queue>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
int n, m, fa[N];
map<pair<int, int> , int> h; 
queue<pair<int, int> > q;
set<int> s[N];
pair<int, int> code(int x, int y) { return make_pair(min(x, y), max(x, y)); }
int find(int x) { return fa[x] == x?x:fa[x] = find(fa[x]); }
int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= (n-1)<<1; ++i)
    {
        int x = read(), y = read();
        if(++h[code(x, y)] == 2) q.push(code(x, y));
        s[x].insert(y), s[y].insert(x);
    }
    while(q.size())
    {
        int x = q.front().first, y = q.front().second; q.pop();
        x = find(x), y = find(y);
        s[x].erase(y), s[y].erase(x);
        if(s[x].size() > s[y].size()) swap(x, y);
        for(set<int>::iterator it = s[x].begin(); it != s[x].end(); ++it)
        {
            h.erase(code(*it, x)), s[*it].erase(x), s[*it].insert(y), s[y].insert(*it);
            if(++h[code(*it, y)] == 2) q.push(code(*it, y));
        }
        s[x].clear(), fa[x] = y;
    }
    for(int i = 1; i <= n; ++i) m += i == find(i);
    if(m == 1) puts("YES"); else puts("NO");
    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 <cstdio>
#include <algorithm>
#include <cctype>
#include <set>
#include <map>
#include <queue>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
int n, m, fa[N];
map<pair<int, int> , int> h; 
queue<pair<int, int> > q;
set<int> s[N];
pair<int, int> code(int x, int y) { return make_pair(min(x, y), max(x, y)); }
int find(int x) { return fa[x] == x?x:fa[x] = find(fa[x]); }
int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= (n-1)<<1; ++i)
    {
        int x = read(), y = read();
        if(++h[code(x, y)] == 2) q.push(code(x, y));
        s[x].insert(y), s[y].insert(x);
    }
    while(q.size())
    {
        int x = q.front().first, y = q.front().second; q.pop();
        x = find(x), y = find(y);
        s[x].erase(y), s[y].erase(x);
        if(s[x].size() > s[y].size()) swap(x, y);
        for(set<int>::iterator it = s[x].begin(); it != s[x].end(); ++it)
        {
            h.erase(code(*it, x)), s[*it].erase(x), s[*it].insert(y), s[y].insert(*it);
            if(++h[code(*it, y)] == 2) q.push(code(*it, y));
        }
        s[x].clear(), fa[x] = y;
    }
    for(int i = 1; i <= n; ++i) m += i == find(i);
    if(m == 1) puts("YES"); else puts("NO");
    return 0;
}
```