#include <bits/stdc++.h>
using namespace std;
long long int n, k, ans, arr[200005][6], siz[200005], km;
vector<long long int> vec[200005];
void dfs(long long int i, long long int pre) {
  siz[i] = 1;
  long long int j, temp, m, r;
  for (j = 0; j < vec[i].size(); j++) {
    if (vec[i][j] == pre) continue;
    dfs(vec[i][j], i);
    temp = siz[vec[i][j]];
    km += temp * (n - temp);
    siz[i] += temp;
    for (m = 0; m < k - 1; m++) ans += (k - (m + 1)) * arr[vec[i][j]][m];
    for (m = 0; m < k; m++) {
      for (r = 0; r < k; r++) {
        if ((r + m + 1) % k == 0) continue;
        ans += (k - ((r + m + 1) % k)) * arr[i][m] * arr[vec[i][j]][r];
      }
    }
    for (m = 0; m < k; m++) {
      arr[i][(m + 1) % k] += arr[vec[i][j]][m];
    }
  }
  arr[i][0]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int a, b, i, j;
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  memset(siz, 0, sizeof(siz));
  memset(arr, 0, sizeof(arr));
  ans = 0;
  dfs(1, -1);
  ans += km;
  ans /= k;
  cout << ans << endl;
  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;
long long int n, k, ans, arr[200005][6], siz[200005], km;
vector<long long int> vec[200005];
void dfs(long long int i, long long int pre) {
  siz[i] = 1;
  long long int j, temp, m, r;
  for (j = 0; j < vec[i].size(); j++) {
    if (vec[i][j] == pre) continue;
    dfs(vec[i][j], i);
    temp = siz[vec[i][j]];
    km += temp * (n - temp);
    siz[i] += temp;
    for (m = 0; m < k - 1; m++) ans += (k - (m + 1)) * arr[vec[i][j]][m];
    for (m = 0; m < k; m++) {
      for (r = 0; r < k; r++) {
        if ((r + m + 1) % k == 0) continue;
        ans += (k - ((r + m + 1) % k)) * arr[i][m] * arr[vec[i][j]][r];
      }
    }
    for (m = 0; m < k; m++) {
      arr[i][(m + 1) % k] += arr[vec[i][j]][m];
    }
  }
  arr[i][0]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int a, b, i, j;
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  memset(siz, 0, sizeof(siz));
  memset(arr, 0, sizeof(arr));
  ans = 0;
  dfs(1, -1);
  ans += km;
  ans /= k;
  cout << ans << endl;
  return 0;
}

```