#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> adj[(int)2e5];
int cnt[(int)2e5];
long long ans = 0;
vector<long long> dfs(int u, int p) {
  vector<vector<long long>> all;
  vector<long long> total(k, 0);
  cnt[u] = 1;
  for (auto &&v : adj[u])
    if (v != p) {
      all.emplace_back(dfs(v, u));
      ans += all.back().back() * (n - cnt[v]);
      rotate(all.back().begin(), all.back().end() - 1, all.back().end());
      cnt[u] += cnt[v];
      for (int i = 0; i < k; ++i) total[i] += all.back()[i];
    }
  long long addAns = 0;
  for (auto &&v : all) {
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        addAns += v[i] * (total[j] - v[j]) * ((i + j + k - 1) / k);
  }
  assert((addAns & 1) == 0);
  ans += addAns >> 1;
  ans += accumulate(total.begin() + 1, total.end(), 0ll);
  ++total[0];
  return total;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them.

Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.

Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.

For a pair of vertices (s, t) we define f(s, t) as the minimum number of jumps Limak needs to get from s to t. Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Input

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) — the number of vertices in the tree and the maximum allowed jump distance respectively.

The next n - 1 lines describe edges in the tree. The i-th of those lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) — the indices on vertices connected with i-th edge.

It's guaranteed that the given edges form a tree.

Output

Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Examples

Input

6 2
1 2
1 3
2 4
2 5
4 6


Output

20


Input

13 3
1 2
3 2
4 2
5 2
3 6
10 6
6 7
6 13
5 8
5 9
9 11
11 12


Output

114


Input

3 5
2 1
3 1


Output

3

Note

In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at most 2. For example, from the vertex 5 he can jump to any of vertices: 1, 2 and 4 (well, he can also jump to the vertex 5 itself).

<image>

There are <image> pairs of vertices (s, t) such that s < t. For 5 of those pairs Limak would need two jumps: (1, 6), (3, 4), (3, 5), (3, 6), (5, 6). For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.

In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t), so the answer is 3·1 = 3.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> adj[(int)2e5];
int cnt[(int)2e5];
long long ans = 0;
vector<long long> dfs(int u, int p) {
  vector<vector<long long>> all;
  vector<long long> total(k, 0);
  cnt[u] = 1;
  for (auto &&v : adj[u])
    if (v != p) {
      all.emplace_back(dfs(v, u));
      ans += all.back().back() * (n - cnt[v]);
      rotate(all.back().begin(), all.back().end() - 1, all.back().end());
      cnt[u] += cnt[v];
      for (int i = 0; i < k; ++i) total[i] += all.back()[i];
    }
  long long addAns = 0;
  for (auto &&v : all) {
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        addAns += v[i] * (total[j] - v[j]) * ((i + j + k - 1) / k);
  }
  assert((addAns & 1) == 0);
  ans += addAns >> 1;
  ans += accumulate(total.begin() + 1, total.end(), 0ll);
  ++total[0];
  return total;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

```