#include <bits/stdc++.h>
using namespace std;
int n, k;
long long res = 0;
vector<int> e[200000];
long long v[200000][6], sum[200000];
long long r[6];
void dfs(int x, int f) {
  for (auto i : e[x]) {
    if (i == f) continue;
    dfs(i, x);
    for (int j = k - 1; j >= 0; j--) v[i][j + 1] = v[i][j];
    v[i][0] = v[i][k];
    for (int j = 0; j < k; j++) v[x][j] += v[i][j];
    sum[x] += sum[i];
  }
  sum[x]++;
  v[x][0]++;
  for (auto i : e[x]) {
    if (i == f) continue;
    for (int ii = 0; ii < k; ii++)
      for (int j = 0; j < k; j++) {
        r[(ii + j) % k] += 1ll * v[i][ii] * (v[x][j] - v[i][j]);
        if (!j) r[(ii + j) % k] += v[i][ii];
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    e[y].push_back(x);
    e[x].push_back(y);
  }
  dfs(0, 0);
  for (int i = 0; i < n; i++) res += 1ll * sum[i] * (n - sum[i]);
  for (int i = 1; i < k; i++) res += 1ll * r[i] * (k - i) / 2;
  cout << res / k;
  return 0;
}

### 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;
int n, k;
long long res = 0;
vector<int> e[200000];
long long v[200000][6], sum[200000];
long long r[6];
void dfs(int x, int f) {
  for (auto i : e[x]) {
    if (i == f) continue;
    dfs(i, x);
    for (int j = k - 1; j >= 0; j--) v[i][j + 1] = v[i][j];
    v[i][0] = v[i][k];
    for (int j = 0; j < k; j++) v[x][j] += v[i][j];
    sum[x] += sum[i];
  }
  sum[x]++;
  v[x][0]++;
  for (auto i : e[x]) {
    if (i == f) continue;
    for (int ii = 0; ii < k; ii++)
      for (int j = 0; j < k; j++) {
        r[(ii + j) % k] += 1ll * v[i][ii] * (v[x][j] - v[i][j]);
        if (!j) r[(ii + j) % k] += v[i][ii];
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    e[y].push_back(x);
    e[x].push_back(y);
  }
  dfs(0, 0);
  for (int i = 0; i < n; i++) res += 1ll * sum[i] * (n - sum[i]);
  for (int i = 1; i < k; i++) res += 1ll * r[i] * (k - i) / 2;
  cout << res / k;
  return 0;
}

```