#include <bits/stdc++.h>
using namespace std;
const int nax = 2e5 + 5;
vector<int> edges[nax];
int count_subtree[nax][5];
int total_subtree[nax];
long long answer;
int n, k;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int a, int par, int depth) {
  count_subtree[a][depth % k] = total_subtree[a] = 1;
  for (int b : edges[a])
    if (b != par) {
      dfs(b, a, depth + 1);
      for (int i = 0; i < k; ++i)
        for (int j = 0; j < k; ++j) {
          int remainder = subtract(i + j, 2 * depth);
          int needs = subtract(k, remainder);
          answer +=
              (long long)needs * count_subtree[a][i] * count_subtree[b][j];
        }
      for (int i = 0; i < k; ++i) count_subtree[a][i] += count_subtree[b][i];
      total_subtree[a] += total_subtree[b];
    }
  answer += (long long)total_subtree[a] * (n - total_subtree[a]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(1, -1, 0);
  assert(answer % k == 0);
  printf("%I64d\n", answer / k);
}

### Prompt

Please create a solution in Cpp 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 int nax = 2e5 + 5;
vector<int> edges[nax];
int count_subtree[nax][5];
int total_subtree[nax];
long long answer;
int n, k;
int subtract(int a, int b) { return ((a - b) % k + k) % k; }
void dfs(int a, int par, int depth) {
  count_subtree[a][depth % k] = total_subtree[a] = 1;
  for (int b : edges[a])
    if (b != par) {
      dfs(b, a, depth + 1);
      for (int i = 0; i < k; ++i)
        for (int j = 0; j < k; ++j) {
          int remainder = subtract(i + j, 2 * depth);
          int needs = subtract(k, remainder);
          answer +=
              (long long)needs * count_subtree[a][i] * count_subtree[b][j];
        }
      for (int i = 0; i < k; ++i) count_subtree[a][i] += count_subtree[b][i];
      total_subtree[a] += total_subtree[b];
    }
  answer += (long long)total_subtree[a] * (n - total_subtree[a]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(1, -1, 0);
  assert(answer % k == 0);
  printf("%I64d\n", answer / k);
}

```