#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int toward, next;
} Edge;
Edge e[300000 * 2];
long long f[300000][5], s[300000][5], sum;
int n, tot, m, chose[300000], fi[300000];
void addedge(int j, int k) {
  ++tot;
  e[tot].toward = k;
  e[tot].next = fi[j];
  fi[j] = tot;
}
void dfs(int x) {
  chose[x] = 1;
  f[x][0] = 0, s[x][0] = 1;
  for (int i = fi[x]; i; i = e[i].next) {
    int too = e[i].toward;
    if (!chose[too]) {
      dfs(too);
      for (int j = 0; j <= m - 1; j++)
        if (s[x][j])
          for (int k = 0; k <= m - 1; k++)
            if (s[too][k]) {
              if (j + k < m)
                sum += f[x][j] * s[too][k] + s[too][k] * s[x][j] +
                       s[x][j] * f[too][k];
              else
                sum += f[x][j] * s[too][k] + s[too][k] * s[x][j] * 2 +
                       s[x][j] * f[too][k];
            }
      f[x][0] += f[too][m - 1] + s[too][m - 1];
      s[x][0] += s[too][m - 1];
      for (int j = 0; j <= m - 2; j++)
        f[x][j + 1] += f[too][j], s[x][j + 1] += s[too][j];
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n - 1; i++) {
    int j, k;
    scanf("%d %d", &j, &k);
    addedge(j, k);
    addedge(k, j);
  }
  dfs(1);
  printf("%I64d\n", sum);
  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;
typedef struct {
  int toward, next;
} Edge;
Edge e[300000 * 2];
long long f[300000][5], s[300000][5], sum;
int n, tot, m, chose[300000], fi[300000];
void addedge(int j, int k) {
  ++tot;
  e[tot].toward = k;
  e[tot].next = fi[j];
  fi[j] = tot;
}
void dfs(int x) {
  chose[x] = 1;
  f[x][0] = 0, s[x][0] = 1;
  for (int i = fi[x]; i; i = e[i].next) {
    int too = e[i].toward;
    if (!chose[too]) {
      dfs(too);
      for (int j = 0; j <= m - 1; j++)
        if (s[x][j])
          for (int k = 0; k <= m - 1; k++)
            if (s[too][k]) {
              if (j + k < m)
                sum += f[x][j] * s[too][k] + s[too][k] * s[x][j] +
                       s[x][j] * f[too][k];
              else
                sum += f[x][j] * s[too][k] + s[too][k] * s[x][j] * 2 +
                       s[x][j] * f[too][k];
            }
      f[x][0] += f[too][m - 1] + s[too][m - 1];
      s[x][0] += s[too][m - 1];
      for (int j = 0; j <= m - 2; j++)
        f[x][j + 1] += f[too][j], s[x][j + 1] += s[too][j];
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n - 1; i++) {
    int j, k;
    scanf("%d %d", &j, &k);
    addedge(j, k);
    addedge(k, j);
  }
  dfs(1);
  printf("%I64d\n", sum);
  return 0;
}

```