#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200042;
vector<int> G[MAXN];
long long Res[MAXN], Cnt[MAXN][5], Sum[MAXN][5];
bool visited[MAXN];
int n, k;
long long dfs(int w) {
  Res[w] = 0;
  visited[w] = true;
  long long tCnt[5], tSum[5];
  for (int i = (0); i <= ((int)k - 1); ++i) {
    tCnt[i] = 0;
    tSum[i] = 0;
  }
  tCnt[0] = 1;
  for (int i = (0); i <= ((int)G[w].size() - 1); ++i) {
    int u = G[w][i];
    if (!visited[u]) {
      dfs(u);
      Res[w] += Res[u];
      for (int l = (0); l <= ((int)k - 1); ++l) {
        for (int r = (0); r <= ((int)k - 1); ++r) {
          long long val =
              tSum[l] * Cnt[u][r] + tCnt[l] * Sum[u][r] + tCnt[l] * Cnt[u][r];
          if (((l + r + 1) % k) != 0) {
            val += (k - ((l + r + 1) % k)) * tCnt[l] * Cnt[u][r];
          }
          Res[w] += (val / k);
        }
      }
      for (int j = (0); j <= ((int)k - 1); ++j) {
        tCnt[(j + 1) % k] += Cnt[u][j];
        tSum[(j + 1) % k] += Sum[u][j] + Cnt[u][j];
      }
    }
  }
  for (int i = (0); i <= ((int)k - 1); ++i) {
    Cnt[w][i] = tCnt[i];
    Sum[w][i] = tSum[i];
  }
  return Res[w];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = (0); i <= ((int)n - 1 - 1); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  cout << dfs(0) << "\n";
  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;
const int MAXN = 200042;
vector<int> G[MAXN];
long long Res[MAXN], Cnt[MAXN][5], Sum[MAXN][5];
bool visited[MAXN];
int n, k;
long long dfs(int w) {
  Res[w] = 0;
  visited[w] = true;
  long long tCnt[5], tSum[5];
  for (int i = (0); i <= ((int)k - 1); ++i) {
    tCnt[i] = 0;
    tSum[i] = 0;
  }
  tCnt[0] = 1;
  for (int i = (0); i <= ((int)G[w].size() - 1); ++i) {
    int u = G[w][i];
    if (!visited[u]) {
      dfs(u);
      Res[w] += Res[u];
      for (int l = (0); l <= ((int)k - 1); ++l) {
        for (int r = (0); r <= ((int)k - 1); ++r) {
          long long val =
              tSum[l] * Cnt[u][r] + tCnt[l] * Sum[u][r] + tCnt[l] * Cnt[u][r];
          if (((l + r + 1) % k) != 0) {
            val += (k - ((l + r + 1) % k)) * tCnt[l] * Cnt[u][r];
          }
          Res[w] += (val / k);
        }
      }
      for (int j = (0); j <= ((int)k - 1); ++j) {
        tCnt[(j + 1) % k] += Cnt[u][j];
        tSum[(j + 1) % k] += Sum[u][j] + Cnt[u][j];
      }
    }
  }
  for (int i = (0); i <= ((int)k - 1); ++i) {
    Cnt[w][i] = tCnt[i];
    Sum[w][i] = tSum[i];
  }
  return Res[w];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = (0); i <= ((int)n - 1 - 1); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  cout << dfs(0) << "\n";
  return 0;
}

```