#include <bits/stdc++.h>
int parent[400000];
int n, k;
int list[400000][2];
int nSucc[400000];
int succ[800000];
int occ[800000];
long long int dist[400000][6];
long long int hops[400000];
long long int subs[400000];
long long int result;
void search(int cur, int par) {
  for (int i = nSucc[cur]; i < nSucc[cur + 1]; ++i) {
    if (succ[i] != par) {
      parent[succ[i]] = cur;
      search(succ[i], cur);
    }
  }
}
void count(int cur) {
  ++dist[cur][0];
  ++subs[cur];
  for (int i = nSucc[cur]; i < nSucc[cur + 1]; ++i) {
    if (succ[i] != parent[cur]) {
      count(succ[i]);
    }
  }
  result += subs[parent[cur]] * (hops[cur]) + subs[cur] * (hops[parent[cur]]);
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < k; ++j) {
      result += dist[parent[cur]][i] * dist[cur][j];
    }
    for (int j = k - i; j < k; ++j) {
      result += dist[parent[cur]][i] * dist[cur][j];
    }
  }
  hops[parent[cur]] += hops[cur];
  subs[parent[cur]] += subs[cur];
  for (int i = 0; i < k; ++i) {
    dist[parent[cur]][i + 1] += dist[cur][i];
  }
  hops[parent[cur]] += dist[parent[cur]][k];
  dist[parent[cur]][0] += dist[parent[cur]][k];
  dist[parent[cur]][k] = 0;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d", &list[i][0], &list[i][1]);
    nSucc[list[i][0]]++;
    nSucc[list[i][1]]++;
    --list[i][0];
    --list[i][1];
  }
  for (int i = 0; i < n + 1; ++i) {
    nSucc[i + 1] += nSucc[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    succ[nSucc[list[i][0]] + (occ[list[i][0]]++)] = list[i][1];
    succ[nSucc[list[i][1]] + (occ[list[i][1]]++)] = list[i][0];
  }
  search(0, -1);
  parent[0] = -1;
  count(0);
  printf("%lld\n", result);
  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>
int parent[400000];
int n, k;
int list[400000][2];
int nSucc[400000];
int succ[800000];
int occ[800000];
long long int dist[400000][6];
long long int hops[400000];
long long int subs[400000];
long long int result;
void search(int cur, int par) {
  for (int i = nSucc[cur]; i < nSucc[cur + 1]; ++i) {
    if (succ[i] != par) {
      parent[succ[i]] = cur;
      search(succ[i], cur);
    }
  }
}
void count(int cur) {
  ++dist[cur][0];
  ++subs[cur];
  for (int i = nSucc[cur]; i < nSucc[cur + 1]; ++i) {
    if (succ[i] != parent[cur]) {
      count(succ[i]);
    }
  }
  result += subs[parent[cur]] * (hops[cur]) + subs[cur] * (hops[parent[cur]]);
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < k; ++j) {
      result += dist[parent[cur]][i] * dist[cur][j];
    }
    for (int j = k - i; j < k; ++j) {
      result += dist[parent[cur]][i] * dist[cur][j];
    }
  }
  hops[parent[cur]] += hops[cur];
  subs[parent[cur]] += subs[cur];
  for (int i = 0; i < k; ++i) {
    dist[parent[cur]][i + 1] += dist[cur][i];
  }
  hops[parent[cur]] += dist[parent[cur]][k];
  dist[parent[cur]][0] += dist[parent[cur]][k];
  dist[parent[cur]][k] = 0;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d", &list[i][0], &list[i][1]);
    nSucc[list[i][0]]++;
    nSucc[list[i][1]]++;
    --list[i][0];
    --list[i][1];
  }
  for (int i = 0; i < n + 1; ++i) {
    nSucc[i + 1] += nSucc[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    succ[nSucc[list[i][0]] + (occ[list[i][0]]++)] = list[i][1];
    succ[nSucc[list[i][1]] + (occ[list[i][1]]++)] = list[i][0];
  }
  search(0, -1);
  parent[0] = -1;
  count(0);
  printf("%lld\n", result);
  return 0;
}

```