#include <cstdio>
#include <map>
#include <queue>
#include <tuple>
#include <utility>

const int N = 100000 + 10;

int n;

int anc[N];

int find(int x) { return anc[x] == x ? x : (anc[x] = find(anc[x])); }

std::map<int, int> adj[N];

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) anc[i] = i;
  std::queue<std::pair<int, int>> q;
  for (int i = 2 * n - 2; i--;) {
    int a, b;
    scanf("%d%d", &a, &b);
    q.emplace(a, b);
    ++adj[a][b];
    ++adj[b][a];
  }
  for (; !q.empty(); q.pop()) {
    int a, b;
    std::tie(a, b) = q.front();
    a = find(a), b = find(b);
    if (a == b || adj[a].find(b) == adj[a].end()) continue;
    if (adj[a][b] < 2) continue;
    if (adj[a][b] > 2) return puts("NO"), 0;
    adj[a].erase(b);
    adj[b].erase(a);
    if (adj[a].size() < adj[b].size()) std::swap(a, b);
    anc[b] = a;
    for (auto it : adj[b]) {
      int c = it.first, d = it.second;
      adj[a][c] += d;
      adj[c][a] += d;
      adj[c].erase(b);
      q.emplace(a, c);
    }
  }
  for (int i = 1; i <= n; ++i) if (find(i) != find(1)) return puts("NO"), 0;
  puts("YES");
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 <map>
#include <queue>
#include <tuple>
#include <utility>

const int N = 100000 + 10;

int n;

int anc[N];

int find(int x) { return anc[x] == x ? x : (anc[x] = find(anc[x])); }

std::map<int, int> adj[N];

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) anc[i] = i;
  std::queue<std::pair<int, int>> q;
  for (int i = 2 * n - 2; i--;) {
    int a, b;
    scanf("%d%d", &a, &b);
    q.emplace(a, b);
    ++adj[a][b];
    ++adj[b][a];
  }
  for (; !q.empty(); q.pop()) {
    int a, b;
    std::tie(a, b) = q.front();
    a = find(a), b = find(b);
    if (a == b || adj[a].find(b) == adj[a].end()) continue;
    if (adj[a][b] < 2) continue;
    if (adj[a][b] > 2) return puts("NO"), 0;
    adj[a].erase(b);
    adj[b].erase(a);
    if (adj[a].size() < adj[b].size()) std::swap(a, b);
    anc[b] = a;
    for (auto it : adj[b]) {
      int c = it.first, d = it.second;
      adj[a][c] += d;
      adj[c][a] += d;
      adj[c].erase(b);
      q.emplace(a, c);
    }
  }
  for (int i = 1; i <= n; ++i) if (find(i) != find(1)) return puts("NO"), 0;
  puts("YES");
  return 0;
}

```