#include <bits/stdc++.h>
using namespace std;
const int N = 200008;
vector<int> a[N];
int n, m, ns[N];
long long ans, b[N], c[N][6];
int dfs(int x, int pre) {
  int i, j, k, son;
  ns[x] = 1;
  b[x] = 0;
  c[x][0] = 1;
  for (i = 0; i < (int)a[x].size(); i++)
    if ((son = a[x][i]) != pre) {
      dfs(son, x);
      ns[x] += ns[son];
      b[x] += ns[son] + b[son];
    }
  for (i = 0; i < (int)a[x].size(); i++)
    if ((son = a[x][i]) != pre) {
      ans += (ns[x] - ns[son]) * (ns[son] + b[son]);
      for (j = 0; j < m; j++)
        for (k = 0; k < m; k++)
          if ((j + k + 1) % m)
            ans += (m - (j + k + 1) % m) * c[x][k] * c[son][j];
      for (j = 0; j < m; j++) c[x][(j + 1) % m] += c[son][j];
    }
  return 0;
}
int main(void) {
  int i, p1, p2;
  scanf("%d%d", &n, &m);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &p1, &p2);
    a[p1].push_back(p2);
    a[p2].push_back(p1);
  }
  dfs(1, 0);
  cout << ans / m;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int N = 200008;
vector<int> a[N];
int n, m, ns[N];
long long ans, b[N], c[N][6];
int dfs(int x, int pre) {
  int i, j, k, son;
  ns[x] = 1;
  b[x] = 0;
  c[x][0] = 1;
  for (i = 0; i < (int)a[x].size(); i++)
    if ((son = a[x][i]) != pre) {
      dfs(son, x);
      ns[x] += ns[son];
      b[x] += ns[son] + b[son];
    }
  for (i = 0; i < (int)a[x].size(); i++)
    if ((son = a[x][i]) != pre) {
      ans += (ns[x] - ns[son]) * (ns[son] + b[son]);
      for (j = 0; j < m; j++)
        for (k = 0; k < m; k++)
          if ((j + k + 1) % m)
            ans += (m - (j + k + 1) % m) * c[x][k] * c[son][j];
      for (j = 0; j < m; j++) c[x][(j + 1) % m] += c[son][j];
    }
  return 0;
}
int main(void) {
  int i, p1, p2;
  scanf("%d%d", &n, &m);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &p1, &p2);
    a[p1].push_back(p2);
    a[p2].push_back(p1);
  }
  dfs(1, 0);
  cout << ans / m;
  return 0;
}

```