#include <bits/stdc++.h>
using namespace std;
long long a, s, d[200003][6], d1[200003][6], f[6], f1[6], g, h, j, k, l, i, n,
    m;
vector<int> v[200003];
void dfs(int idx, int par) {
  for (auto it : v[idx]) {
    if (it == par) continue;
    dfs(it, idx);
    d[idx][0]++;
    d1[idx][0]++;
    for (int i = 1; i < m; i++) {
      d[idx][i] += d[it][i - 1];
      d1[idx][i] += d1[it][i - 1];
    }
    d[idx][0] += d[it][m - 1] + d1[it][m - 1];
    d1[idx][0] += d1[it][m - 1];
  }
  for (int i = 0; i < m; i++) {
    k += 2 * d[idx][i];
  }
  d[idx][0]--;
  d1[idx][0]--;
  for (auto it : v[idx]) {
    if (it == par) continue;
    for (int i = 1; i < m; i++) {
      f[i] = d[it][i - 1];
      f1[i] = d1[it][i - 1];
      d[idx][i] -= f[i];
      d1[idx][i] -= f1[i];
    }
    f[0] = d[it][m - 1] + d1[it][m - 1];
    f1[0] = d1[it][m - 1];
    d[idx][0] -= f[0];
    d1[idx][0] -= f1[0];
    f[0]++;
    f1[0]++;
    for (int i = 0; i < m; i++) {
      if (!d1[idx][i]) continue;
      for (int a = 0; a < m; a++) {
        if (!f1[a]) continue;
        if (i + a + 2 <= m)
          k += (d[idx][i] - d1[idx][i]) * f1[a] + (f[a] - f1[a]) * d1[idx][i] +
               f1[a] * d1[idx][i];
        else {
          k += d[idx][i] * f1[a] + f[a] * d1[idx][i];
        }
      }
    }
    f[0]--;
    f1[0]--;
    for (int i = 0; i < m; i++) {
      d[idx][i] += f[i];
      d1[idx][i] += f1[i];
    }
  }
  d[idx][0]++;
  d1[idx][0]++;
}
int main() {
  cin >> n >> m;
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &s);
    v[a].push_back(s);
    v[s].push_back(a);
  }
  dfs(1, 0);
  cout << k / 2;
}

### Prompt

Your task is to create 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;
long long a, s, d[200003][6], d1[200003][6], f[6], f1[6], g, h, j, k, l, i, n,
    m;
vector<int> v[200003];
void dfs(int idx, int par) {
  for (auto it : v[idx]) {
    if (it == par) continue;
    dfs(it, idx);
    d[idx][0]++;
    d1[idx][0]++;
    for (int i = 1; i < m; i++) {
      d[idx][i] += d[it][i - 1];
      d1[idx][i] += d1[it][i - 1];
    }
    d[idx][0] += d[it][m - 1] + d1[it][m - 1];
    d1[idx][0] += d1[it][m - 1];
  }
  for (int i = 0; i < m; i++) {
    k += 2 * d[idx][i];
  }
  d[idx][0]--;
  d1[idx][0]--;
  for (auto it : v[idx]) {
    if (it == par) continue;
    for (int i = 1; i < m; i++) {
      f[i] = d[it][i - 1];
      f1[i] = d1[it][i - 1];
      d[idx][i] -= f[i];
      d1[idx][i] -= f1[i];
    }
    f[0] = d[it][m - 1] + d1[it][m - 1];
    f1[0] = d1[it][m - 1];
    d[idx][0] -= f[0];
    d1[idx][0] -= f1[0];
    f[0]++;
    f1[0]++;
    for (int i = 0; i < m; i++) {
      if (!d1[idx][i]) continue;
      for (int a = 0; a < m; a++) {
        if (!f1[a]) continue;
        if (i + a + 2 <= m)
          k += (d[idx][i] - d1[idx][i]) * f1[a] + (f[a] - f1[a]) * d1[idx][i] +
               f1[a] * d1[idx][i];
        else {
          k += d[idx][i] * f1[a] + f[a] * d1[idx][i];
        }
      }
    }
    f[0]--;
    f1[0]--;
    for (int i = 0; i < m; i++) {
      d[idx][i] += f[i];
      d1[idx][i] += f1[i];
    }
  }
  d[idx][0]++;
  d1[idx][0]++;
}
int main() {
  cin >> n >> m;
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &s);
    v[a].push_back(s);
    v[s].push_back(a);
  }
  dfs(1, 0);
  cout << k / 2;
}

```