#include <bits/stdc++.h>
using namespace std;
long long s[220000][5], elem[220000][5], auxs[220000][5], auxe[220000][5];
long long k;
vector<long long> grafo[220000];
long long tot = 0;
void dfs(long long num, long long padre) {
  elem[num][0] = 1;
  long long hijo;
  for (long long i = 0; i < grafo[num].size(); ++i) {
    hijo = grafo[num][i];
    if (hijo == padre) continue;
    dfs(hijo, num);
    auxs[hijo][1 % k] += elem[hijo][0];
    for (long long j = 0; j < k; j++) {
      auxs[hijo][(j + 1) % k] += s[hijo][j];
      auxe[hijo][(j + 1) % k] += elem[hijo][j];
    }
    for (long long j = 0; j < k; j++) {
      for (long long l = 0; l < k; l++) {
        if (elem[num][j] > 0 && auxe[hijo][l] > 0) {
          if (l + j > k || l == 0 || j == 0) {
            tot += s[num][j] * auxe[hijo][l] + auxs[hijo][l] * elem[num][j];
          } else {
            tot += s[num][j] * auxe[hijo][l] + auxs[hijo][l] * elem[num][j] -
                   auxe[hijo][l] * elem[num][j];
          }
        }
      }
    }
    for (long long j = 0; j < k; j++) {
      elem[num][j] += auxe[hijo][j];
      s[num][j] += auxs[hijo][j];
    }
  }
}
void clr() {
  for (long long i = 0; i < 220000; ++i) {
    for (long long j = 0; j < 5; j++) {
      s[i][j] = elem[i][j] = auxe[i][j] = auxs[i][j] = 0;
    }
  }
}
int main() {
  clr();
  long long n;
  cin >> n >> k;
  long long iz, der;
  for (long long i = 0; i < n - 1; ++i) {
    cin >> iz >> der;
    grafo[iz].push_back(der);
    grafo[der].push_back(iz);
  }
  dfs(1, 1);
  cout << tot << endl;
}

### Prompt

In Cpp, your task is to solve 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;
long long s[220000][5], elem[220000][5], auxs[220000][5], auxe[220000][5];
long long k;
vector<long long> grafo[220000];
long long tot = 0;
void dfs(long long num, long long padre) {
  elem[num][0] = 1;
  long long hijo;
  for (long long i = 0; i < grafo[num].size(); ++i) {
    hijo = grafo[num][i];
    if (hijo == padre) continue;
    dfs(hijo, num);
    auxs[hijo][1 % k] += elem[hijo][0];
    for (long long j = 0; j < k; j++) {
      auxs[hijo][(j + 1) % k] += s[hijo][j];
      auxe[hijo][(j + 1) % k] += elem[hijo][j];
    }
    for (long long j = 0; j < k; j++) {
      for (long long l = 0; l < k; l++) {
        if (elem[num][j] > 0 && auxe[hijo][l] > 0) {
          if (l + j > k || l == 0 || j == 0) {
            tot += s[num][j] * auxe[hijo][l] + auxs[hijo][l] * elem[num][j];
          } else {
            tot += s[num][j] * auxe[hijo][l] + auxs[hijo][l] * elem[num][j] -
                   auxe[hijo][l] * elem[num][j];
          }
        }
      }
    }
    for (long long j = 0; j < k; j++) {
      elem[num][j] += auxe[hijo][j];
      s[num][j] += auxs[hijo][j];
    }
  }
}
void clr() {
  for (long long i = 0; i < 220000; ++i) {
    for (long long j = 0; j < 5; j++) {
      s[i][j] = elem[i][j] = auxe[i][j] = auxs[i][j] = 0;
    }
  }
}
int main() {
  clr();
  long long n;
  cin >> n >> k;
  long long iz, der;
  for (long long i = 0; i < n - 1; ++i) {
    cin >> iz >> der;
    grafo[iz].push_back(der);
    grafo[der].push_back(iz);
  }
  dfs(1, 1);
  cout << tot << endl;
}

```