#include <bits/stdc++.h>
using namespace std;
const int SO = 2e5 + 5;
int n, k;
long long f[SO][6], g[SO][6], b[SO][6], kq;
vector<int> ke[SO];
long long fix(long long so) { return ((so + k) % k); }
long long fixx(long long so) {
  long long b = 0;
  return max(so, b);
}
void DFS(int u, int p) {
  int dem = 0;
  long long tf[6], tg[6];
  for (int c = 0; c < ke[u].size(); c++) {
    int v = ke[u][c];
    if (v != p) {
      DFS(v, u);
      dem++;
      for (int i = 0; i < k; i++) {
        f[u][i] += f[v][fix(i - 1)];
        tf[i] += f[v][fix(i - 1)];
        g[u][i] += g[v][fix(i - 1)];
        b[v][i] = g[v][fix(i - 1)];
      }
      g[u][0] += f[v][k - 1];
      b[v][0] += f[v][k - 1];
    }
  }
  long long kqq = 0;
  if (dem > 1)
    for (int c = 0; c < ke[u].size(); c++) {
      int v = ke[u][c];
      if (v != p) {
        for (int i = 0; i < k; i++)
          for (int j = 0; j < k; j++)
            kqq += f[v][fix(i - 1)] * (g[u][j] - b[v][j]) +
                   b[v][i] * (f[u][j] - f[v][fix(j - 1)]) +
                   f[v][fix(i - 1)] * (f[u][j] - f[v][fix(j - 1)]) *
                       fixx((i + j - 1) / k + 1 - ((i + j) == 0));
      }
    }
  kq += kqq / 2;
  for (int i = 0; i < k; i++) {
    kq += g[u][i];
    if (i > 0) kq += f[u][i];
  }
  f[u][0]++;
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ke[x].push_back(y);
    ke[y].push_back(x);
  }
  DFS(1, 0);
  printf("%I64d", kq);
  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 int SO = 2e5 + 5;
int n, k;
long long f[SO][6], g[SO][6], b[SO][6], kq;
vector<int> ke[SO];
long long fix(long long so) { return ((so + k) % k); }
long long fixx(long long so) {
  long long b = 0;
  return max(so, b);
}
void DFS(int u, int p) {
  int dem = 0;
  long long tf[6], tg[6];
  for (int c = 0; c < ke[u].size(); c++) {
    int v = ke[u][c];
    if (v != p) {
      DFS(v, u);
      dem++;
      for (int i = 0; i < k; i++) {
        f[u][i] += f[v][fix(i - 1)];
        tf[i] += f[v][fix(i - 1)];
        g[u][i] += g[v][fix(i - 1)];
        b[v][i] = g[v][fix(i - 1)];
      }
      g[u][0] += f[v][k - 1];
      b[v][0] += f[v][k - 1];
    }
  }
  long long kqq = 0;
  if (dem > 1)
    for (int c = 0; c < ke[u].size(); c++) {
      int v = ke[u][c];
      if (v != p) {
        for (int i = 0; i < k; i++)
          for (int j = 0; j < k; j++)
            kqq += f[v][fix(i - 1)] * (g[u][j] - b[v][j]) +
                   b[v][i] * (f[u][j] - f[v][fix(j - 1)]) +
                   f[v][fix(i - 1)] * (f[u][j] - f[v][fix(j - 1)]) *
                       fixx((i + j - 1) / k + 1 - ((i + j) == 0));
      }
    }
  kq += kqq / 2;
  for (int i = 0; i < k; i++) {
    kq += g[u][i];
    if (i > 0) kq += f[u][i];
  }
  f[u][0]++;
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ke[x].push_back(y);
    ke[y].push_back(x);
  }
  DFS(1, 0);
  printf("%I64d", kq);
  return 0;
}

```