#include <bits/stdc++.h>
using namespace std;
const double eps = 0.000000000000001;
const int N = 200010;
vector<int> gr[N];
long long way[N][6], sum[N][6];
int n, k;
long long ans = 0;
void dfs(int c, int p) {
  int i, z = gr[c].size(), a, b;
  int x;
  way[c][0] = 1;
  for (i = 0; i < z; i++) {
    if (gr[c][i] == p) continue;
    int v = gr[c][i];
    dfs(v, c);
    for (a = 0; a < k; a++) {
      for (b = 0; b < k; b++) {
        ans += sum[v][b] * way[c][a];
        ans += (way[v][b] * sum[c][a]);
        x = (a + b + 1) / k + ((a + b + 1) == k ? 0 : 1);
        ans += way[c][a] * x * way[v][b];
      }
    }
    for (a = 0; a < k; a++) {
      x = (a == k - 1 ? 0 : a + 1);
      way[c][x] += way[v][a];
      sum[c][x] += sum[v][a] + (a == k - 1 ? 1 : 0) * way[v][a];
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  int i, j;
  for (i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
const double eps = 0.000000000000001;
const int N = 200010;
vector<int> gr[N];
long long way[N][6], sum[N][6];
int n, k;
long long ans = 0;
void dfs(int c, int p) {
  int i, z = gr[c].size(), a, b;
  int x;
  way[c][0] = 1;
  for (i = 0; i < z; i++) {
    if (gr[c][i] == p) continue;
    int v = gr[c][i];
    dfs(v, c);
    for (a = 0; a < k; a++) {
      for (b = 0; b < k; b++) {
        ans += sum[v][b] * way[c][a];
        ans += (way[v][b] * sum[c][a]);
        x = (a + b + 1) / k + ((a + b + 1) == k ? 0 : 1);
        ans += way[c][a] * x * way[v][b];
      }
    }
    for (a = 0; a < k; a++) {
      x = (a == k - 1 ? 0 : a + 1);
      way[c][x] += way[v][a];
      sum[c][x] += sum[v][a] + (a == k - 1 ? 1 : 0) * way[v][a];
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  int i, j;
  for (i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```