#include <bits/stdc++.h>
int eCnt, eLast[200005], ePrev[400005], eTar[400005];
void AddEdge(int u, int v) {
  ePrev[++eCnt] = eLast[u];
  eTar[eCnt] = v;
  eLast[u] = eCnt;
}
int m;
long long ans, f[200005], g[200005], h[200005][5], t[200005];
void Solve(int now, int far) {
  int i, j, k, next;
  f[now] = 1;
  g[now] = 0;
  h[now][0] = 1;
  for (i = eLast[now]; i; i = ePrev[i]) {
    if ((next = eTar[i]) == far) continue;
    Solve(next, now);
    ans += f[now] * g[next] + g[now] * f[next];
    for (j = 0; j < m; ++j)
      for (k = 0; k < m; ++k)
        ans += (1 + (j + k + 1 > m)) * h[now][j] * h[next][k];
    f[now] += f[next];
    g[now] += g[next] + h[next][m - 1];
    for (j = 0; j < m; ++j) h[now][(j + 1) % m] += h[next][j];
  }
}
int main(int argc, char *argv[]) {
  int n, i, u, v;
  scanf("%d%d", &n, &m);
  for (i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    AddEdge(u, v);
    AddEdge(v, u);
  }
  Solve(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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>
int eCnt, eLast[200005], ePrev[400005], eTar[400005];
void AddEdge(int u, int v) {
  ePrev[++eCnt] = eLast[u];
  eTar[eCnt] = v;
  eLast[u] = eCnt;
}
int m;
long long ans, f[200005], g[200005], h[200005][5], t[200005];
void Solve(int now, int far) {
  int i, j, k, next;
  f[now] = 1;
  g[now] = 0;
  h[now][0] = 1;
  for (i = eLast[now]; i; i = ePrev[i]) {
    if ((next = eTar[i]) == far) continue;
    Solve(next, now);
    ans += f[now] * g[next] + g[now] * f[next];
    for (j = 0; j < m; ++j)
      for (k = 0; k < m; ++k)
        ans += (1 + (j + k + 1 > m)) * h[now][j] * h[next][k];
    f[now] += f[next];
    g[now] += g[next] + h[next][m - 1];
    for (j = 0; j < m; ++j) h[now][(j + 1) % m] += h[next][j];
  }
}
int main(int argc, char *argv[]) {
  int n, i, u, v;
  scanf("%d%d", &n, &m);
  for (i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    AddEdge(u, v);
    AddEdge(v, u);
  }
  Solve(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

```