#include <bits/stdc++.h>
const int N = 2e5 + 1;
long long t[5], g[N][5], r;
int i, j, x, y, n, c, z, v, s[5], fa[N], en[N * 2], nxt[N * 2], siz[N][5];
void dfs(int u, int p) {
  for (int i = fa[u]; i; i = nxt[i])
    if (en[i] ^ p) dfs(en[i], u);
  for (memset(t, 0, 8 * c), memset(s, 0, 4 * c), *siz[u] = 1, i = fa[u]; i;
       i = nxt[i])
    if ((v = en[i]) ^ p)
      for (x = 0; x < c; t[x] += g[v][x], s[x] += siz[v][x],
          siz[u][(x + 1) % c] += siz[v][x],
          g[u][(x + 1) % c] += g[v][x] + !x * siz[v][x], ++x)
        for (y = 0; y < c; ++y)
          z = (x - c) % c + (y - c) % c + 2,
          r += t[x] * siz[v][y] + g[v][y] * s[x] +
               1ll * (z < 0 ? z / c : (z + c - 1) / c) * siz[v][y] * s[x];
  for (x = 0; x < c; ++x) r += g[u][x];
}
int main() {
  for (scanf("%d%d", &n, &c); ++x < n;) {
    scanf("%d%d", &i, &j);
    en[++y] = j, nxt[y] = fa[i], fa[i] = y;
    en[++y] = i, nxt[y] = fa[j], fa[j] = y;
  }
  dfs(1, i = 0), printf("%I64d", r);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
const int N = 2e5 + 1;
long long t[5], g[N][5], r;
int i, j, x, y, n, c, z, v, s[5], fa[N], en[N * 2], nxt[N * 2], siz[N][5];
void dfs(int u, int p) {
  for (int i = fa[u]; i; i = nxt[i])
    if (en[i] ^ p) dfs(en[i], u);
  for (memset(t, 0, 8 * c), memset(s, 0, 4 * c), *siz[u] = 1, i = fa[u]; i;
       i = nxt[i])
    if ((v = en[i]) ^ p)
      for (x = 0; x < c; t[x] += g[v][x], s[x] += siz[v][x],
          siz[u][(x + 1) % c] += siz[v][x],
          g[u][(x + 1) % c] += g[v][x] + !x * siz[v][x], ++x)
        for (y = 0; y < c; ++y)
          z = (x - c) % c + (y - c) % c + 2,
          r += t[x] * siz[v][y] + g[v][y] * s[x] +
               1ll * (z < 0 ? z / c : (z + c - 1) / c) * siz[v][y] * s[x];
  for (x = 0; x < c; ++x) r += g[u][x];
}
int main() {
  for (scanf("%d%d", &n, &c); ++x < n;) {
    scanf("%d%d", &i, &j);
    en[++y] = j, nxt[y] = fa[i], fa[i] = y;
    en[++y] = i, nxt[y] = fa[j], fa[j] = y;
  }
  dfs(1, i = 0), printf("%I64d", r);
}

```