#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <cstdlib>
using namespace std;

const int N = 1e5 + 50;

int n;
int fa[N], a[N], b[N], top;
map <int, int> e[N];

inline int getf(int x) {
  return fa[x] ? fa[x] = getf(fa[x]) : x;
}

inline void Merge(int x, int y) {
  x = getf(x); y = getf(y);
  if (x == y) { puts("NO"); exit(0); }
  if (e[x].size() > e[y].size()) swap(x, y);
  fa[x] = y;
  int t;
  for (auto i : e[x]) {
    t = i.first;
    e[t].erase(x);
    if (e[y].count(t)) {
      if (e[y][t] != i.second) {
        a[++ top] = y; b[top] = t;
        e[y].erase(t);
        e[t].erase(y);
      }
      else {
        puts("NO"); exit(0);
      }
    }
    else {
      e[y][t] = i.second;
      e[t][y] = i.second;
    }
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i ++) {
    scanf("%d%d", &u, &v);
    e[u][v] = 1; e[v][u] = 1;
  }
  for (int i = 1, x, y; i < n; i ++) {
    scanf("%d%d", &x, &y);
    if (e[x].count(y)) {
      if (e[x][y] == 2) return 0 * puts("NO");
      e[x].erase(y);
      e[y].erase(x);
      a[++ top] = x; b[top] = y;
    }
    else {
      e[x][y] = 2; e[y][x] = 2;
    }
  }
  for (int i = 1; i <= top; i ++) Merge(a[i], b[i]);
  puts(top + 1 == n ? "YES" : "NO");
  return 0;
}
### Prompt

Generate 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 <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <cstdlib>
using namespace std;

const int N = 1e5 + 50;

int n;
int fa[N], a[N], b[N], top;
map <int, int> e[N];

inline int getf(int x) {
  return fa[x] ? fa[x] = getf(fa[x]) : x;
}

inline void Merge(int x, int y) {
  x = getf(x); y = getf(y);
  if (x == y) { puts("NO"); exit(0); }
  if (e[x].size() > e[y].size()) swap(x, y);
  fa[x] = y;
  int t;
  for (auto i : e[x]) {
    t = i.first;
    e[t].erase(x);
    if (e[y].count(t)) {
      if (e[y][t] != i.second) {
        a[++ top] = y; b[top] = t;
        e[y].erase(t);
        e[t].erase(y);
      }
      else {
        puts("NO"); exit(0);
      }
    }
    else {
      e[y][t] = i.second;
      e[t][y] = i.second;
    }
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i ++) {
    scanf("%d%d", &u, &v);
    e[u][v] = 1; e[v][u] = 1;
  }
  for (int i = 1, x, y; i < n; i ++) {
    scanf("%d%d", &x, &y);
    if (e[x].count(y)) {
      if (e[x][y] == 2) return 0 * puts("NO");
      e[x].erase(y);
      e[y].erase(x);
      a[++ top] = x; b[top] = y;
    }
    else {
      e[x][y] = 2; e[y][x] = 2;
    }
  }
  for (int i = 1; i <= top; i ++) Merge(a[i], b[i]);
  puts(top + 1 == n ? "YES" : "NO");
  return 0;
}
```