#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
std::vector<int> G[N];
int path[N][6], n, k;
long long int s = 0, f = 0;
int dfs(int v, int p = -1) {
  for (int j = 0; j <= k; j++) path[v][j] = 0;
  int sz = 1, val;
  int pth[] = {0, 0, 0, 0, 0, 0};
  bool first = true;
  for (auto u : G[v]) {
    if (u == p) continue;
    val = dfs(u, v), sz += val, s += 1ll * val * (n - val);
    rotate(path[u], path[u] + k - 1, path[u] + k);
    path[u][1]++;
    for (int j = 0; j < k; j++) {
      path[v][j] += path[u][j];
      f += path[u][j] * (j ? k - j : 0);
    }
    if (!first) {
      for (int j = 0; j < k; j++) {
        for (int x = 0; x < k; x++) {
          int len = (j + x) % k;
          f += 1ll * pth[j] * path[u][x] * (len ? k - len : 0);
        }
      }
    } else {
      first = false;
    }
    for (int j = 0; j < k; j++) pth[j] += path[u][j];
  }
  return sz;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b), G[b].push_back(a);
  }
  dfs(1);
  assert((s + f) % k == 0);
  cout << (s + f) / k;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 2e5 + 5;
std::vector<int> G[N];
int path[N][6], n, k;
long long int s = 0, f = 0;
int dfs(int v, int p = -1) {
  for (int j = 0; j <= k; j++) path[v][j] = 0;
  int sz = 1, val;
  int pth[] = {0, 0, 0, 0, 0, 0};
  bool first = true;
  for (auto u : G[v]) {
    if (u == p) continue;
    val = dfs(u, v), sz += val, s += 1ll * val * (n - val);
    rotate(path[u], path[u] + k - 1, path[u] + k);
    path[u][1]++;
    for (int j = 0; j < k; j++) {
      path[v][j] += path[u][j];
      f += path[u][j] * (j ? k - j : 0);
    }
    if (!first) {
      for (int j = 0; j < k; j++) {
        for (int x = 0; x < k; x++) {
          int len = (j + x) % k;
          f += 1ll * pth[j] * path[u][x] * (len ? k - len : 0);
        }
      }
    } else {
      first = false;
    }
    for (int j = 0; j < k; j++) pth[j] += path[u][j];
  }
  return sz;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b), G[b].push_back(a);
  }
  dfs(1);
  assert((s + f) % k == 0);
  cout << (s + f) / k;
}

```