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.
#include <bits/stdc++.h>
using namespace std;
const int N = 200200, D = 5;
int i, j, k, n, m, ch, x, y, En;
int h[N], f[N][D];
long long ans, g[N];
struct edge {
  int s, n;
} E[N << 1];
void R(int &x) {
  x = 0;
  ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
}
void W(long long x) {
  if (x >= 10) W(x / 10);
  putchar(x % 10 + '0');
}
void E_add(int x, int y) {
  E[++En].s = y;
  E[En].n = h[x];
  h[x] = En;
  E[++En].s = x;
  E[En].n = h[y];
  h[y] = En;
}
void dfs1(int x, int F) {
  f[x][0]++;
  for (int k = h[x]; k; k = E[k].n)
    if (E[k].s != F) {
      dfs1(E[k].s, x);
      for (int i = 0; i < m; i++) f[x][(i + 1) % m] += f[E[k].s][i];
      g[x] += g[E[k].s] + f[E[k].s][0];
    }
}
void dfs2(int x, int Fa, int *F, long long G) {
  int tf[D];
  for (int i = 0; i < m; i++) tf[(i + 1) % m] = F[i];
  ans += G + g[x];
  G += F[0];
  for (int i = 0; i < m; i++) tf[(i + 1) % m] += f[x][i];
  G += g[x] + f[x][0];
  for (int k = h[x]; k; k = E[k].n)
    if (E[k].s != Fa) {
      for (int i = 0; i < m; i++) tf[(i + 2) % m] -= f[E[k].s][i];
      G -= g[E[k].s] + f[E[k].s][0] + f[E[k].s][m - 1];
      dfs2(E[k].s, x, tf, G);
      for (int i = 0; i < m; i++) tf[(i + 2) % m] += f[E[k].s][i];
      G += g[E[k].s] + f[E[k].s][0] + f[E[k].s][m - 1];
    }
}
int main() {
  R(n);
  R(m);
  for (i = 1; i < n; i++) {
    R(x);
    R(y);
    E_add(x, y);
  }
  dfs1(1, 0);
  int tmp[D];
  for (i = 0; i < m; i++) tmp[i] = 0;
  dfs2(1, 0, tmp, 0);
  W(ans / 2);
  puts("");
}
