#include <bits/stdc++.h>
const int ms = 200200;
int n, k;
std::vector<int> edges[ms];
long long dists[ms], distt[ms];
long long paths[ms][5];
long long tot[ms];
long long nxt[5];
long long top[ms][5];
long long wtf[5];
long long tot_dist = 0;
void dfs(int on, int par = -1) {
  paths[on][0]++;
  top[on][0]++;
  tot[on]++;
  for (int i = 0; i < edges[on].size(); i++) {
    int to = edges[on][i];
    if (to == par) continue;
    dfs(to, on);
    dists[on] +=
        ((distt[to]) * (tot[on])) + ((distt[on]) * tot[to]) + tot[to] * tot[on];
    distt[on] += distt[to] + tot[to];
    tot[on] += tot[to];
    for (int j = 0; j < k; j++) nxt[j] = paths[on][j];
    for (int j = 0; j < k; j++) {
      for (int jj = 0; jj < k; jj++) {
        nxt[(j + jj + 1) % k] += top[on][j] * top[to][jj];
      }
    }
    for (int j = 0; j < k; j++) paths[on][j] = nxt[j];
    for (int j = 0; j < k; j++) top[on][(j + 1) % k] += top[to][j];
  }
  for (int i = 0; i < k; i++) wtf[i] += paths[on][i];
  tot_dist += dists[on];
}
int main() {
  std::cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%i %i", &u, &v);
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  dfs(1);
  long long ans = tot_dist;
  for (int i = 1; i < k; i++) {
    ans += wtf[i] * (k - i);
  }
  ans /= k;
  std::cout << ans << '\n';
}

### Prompt

Generate 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>
const int ms = 200200;
int n, k;
std::vector<int> edges[ms];
long long dists[ms], distt[ms];
long long paths[ms][5];
long long tot[ms];
long long nxt[5];
long long top[ms][5];
long long wtf[5];
long long tot_dist = 0;
void dfs(int on, int par = -1) {
  paths[on][0]++;
  top[on][0]++;
  tot[on]++;
  for (int i = 0; i < edges[on].size(); i++) {
    int to = edges[on][i];
    if (to == par) continue;
    dfs(to, on);
    dists[on] +=
        ((distt[to]) * (tot[on])) + ((distt[on]) * tot[to]) + tot[to] * tot[on];
    distt[on] += distt[to] + tot[to];
    tot[on] += tot[to];
    for (int j = 0; j < k; j++) nxt[j] = paths[on][j];
    for (int j = 0; j < k; j++) {
      for (int jj = 0; jj < k; jj++) {
        nxt[(j + jj + 1) % k] += top[on][j] * top[to][jj];
      }
    }
    for (int j = 0; j < k; j++) paths[on][j] = nxt[j];
    for (int j = 0; j < k; j++) top[on][(j + 1) % k] += top[to][j];
  }
  for (int i = 0; i < k; i++) wtf[i] += paths[on][i];
  tot_dist += dists[on];
}
int main() {
  std::cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%i %i", &u, &v);
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  dfs(1);
  long long ans = tot_dist;
  for (int i = 1; i < k; i++) {
    ans += wtf[i] * (k - i);
  }
  ans /= k;
  std::cout << ans << '\n';
}

```