#include <bits/stdc++.h>
using namespace std;
int N, K;
long long res = 0;
vector<int> path[200001];
int kagi[200001][5];
long long jumps[200001][5];
void dfs(int pos, int par) {
  int X = path[pos].size();
  int berapa[5];
  long long tmp_jumps[5];
  memset(berapa, 0, sizeof(berapa));
  memset(tmp_jumps, 0, sizeof(tmp_jumps));
  for (int i = 0; i < X; i++) {
    int u = path[pos][i];
    if (u == par) continue;
    dfs(u, pos);
    for (int j = 0; j < K; j++) {
      for (int k = 0; k < K; k++) {
        long long wow = (long long)tmp_jumps[j] * kagi[u][k] +
                        (long long)berapa[j] * jumps[u][k];
        if (j + k > 0) wow += (long long)berapa[j] * kagi[u][k];
        if (j + k > K) wow += (long long)berapa[j] * kagi[u][k];
        res += wow;
      }
    }
    long long bzz = 0;
    for (int j = 0; j < K; j++) {
      if (j)
        bzz += (kagi[u][j] + jumps[u][j]);
      else
        bzz += jumps[u][j];
      berapa[j] += kagi[u][j];
      tmp_jumps[j] += jumps[u][j];
      kagi[pos][j] += kagi[u][j];
      jumps[pos][j] += jumps[u][j];
    }
    res += bzz;
  }
  kagi[pos][0]++;
  int tmp = kagi[pos][K - 1];
  long long tmp2 = jumps[pos][K - 1];
  for (int i = K - 1; i > 0; i--) {
    kagi[pos][i] = kagi[pos][i - 1];
    jumps[pos][i] = jumps[pos][i - 1];
  }
  kagi[pos][0] = tmp;
  jumps[pos][0] = tmp2;
  jumps[pos][0] += kagi[pos][0];
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i < N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  dfs(0, -1);
  cout << res << endl;
}

### Prompt

Please formulate 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;
long long res = 0;
vector<int> path[200001];
int kagi[200001][5];
long long jumps[200001][5];
void dfs(int pos, int par) {
  int X = path[pos].size();
  int berapa[5];
  long long tmp_jumps[5];
  memset(berapa, 0, sizeof(berapa));
  memset(tmp_jumps, 0, sizeof(tmp_jumps));
  for (int i = 0; i < X; i++) {
    int u = path[pos][i];
    if (u == par) continue;
    dfs(u, pos);
    for (int j = 0; j < K; j++) {
      for (int k = 0; k < K; k++) {
        long long wow = (long long)tmp_jumps[j] * kagi[u][k] +
                        (long long)berapa[j] * jumps[u][k];
        if (j + k > 0) wow += (long long)berapa[j] * kagi[u][k];
        if (j + k > K) wow += (long long)berapa[j] * kagi[u][k];
        res += wow;
      }
    }
    long long bzz = 0;
    for (int j = 0; j < K; j++) {
      if (j)
        bzz += (kagi[u][j] + jumps[u][j]);
      else
        bzz += jumps[u][j];
      berapa[j] += kagi[u][j];
      tmp_jumps[j] += jumps[u][j];
      kagi[pos][j] += kagi[u][j];
      jumps[pos][j] += jumps[u][j];
    }
    res += bzz;
  }
  kagi[pos][0]++;
  int tmp = kagi[pos][K - 1];
  long long tmp2 = jumps[pos][K - 1];
  for (int i = K - 1; i > 0; i--) {
    kagi[pos][i] = kagi[pos][i - 1];
    jumps[pos][i] = jumps[pos][i - 1];
  }
  kagi[pos][0] = tmp;
  jumps[pos][0] = tmp2;
  jumps[pos][0] += kagi[pos][0];
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i < N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  dfs(0, -1);
  cout << res << endl;
}

```