#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 5;
int cc[N][M], sz[N], k;
long long ret[N], sum = 0;
vector<int> adj[N];
long long dfs(int x, int p) {
  int t, v, j, u, i;
  sz[x] = 1;
  long long R = 0, temp;
  for (i = 0; i < adj[x].size(); ++i) {
    u = adj[x][i];
    if (u != p) {
      temp = dfs(u, x);
      sum += sz[u] * R;
      R += temp;
      sum += temp * sz[x];
      for (j = 0; j < k; ++j)
        for (t = 0; t < k; ++t) {
          if (j + t + 2 <= k) sum -= (long long)cc[x][j] * cc[u][t];
        }
      for (j = 0; j < k; ++j) cc[x][j] += cc[u][j];
      sz[x] += sz[u];
    }
  }
  u = ++cc[x][k - 1];
  for (i = k - 2; i >= 0; --i) cc[x][i + 1] = cc[x][i];
  cc[x][0] = u;
  ret[x] = R + u;
  return ret[x];
}
int main() {
  int n, i, j, u, v, a, b;
  cin >> n >> k;
  for (i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  cout << sum << endl;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 2e5 + 5, M = 5;
int cc[N][M], sz[N], k;
long long ret[N], sum = 0;
vector<int> adj[N];
long long dfs(int x, int p) {
  int t, v, j, u, i;
  sz[x] = 1;
  long long R = 0, temp;
  for (i = 0; i < adj[x].size(); ++i) {
    u = adj[x][i];
    if (u != p) {
      temp = dfs(u, x);
      sum += sz[u] * R;
      R += temp;
      sum += temp * sz[x];
      for (j = 0; j < k; ++j)
        for (t = 0; t < k; ++t) {
          if (j + t + 2 <= k) sum -= (long long)cc[x][j] * cc[u][t];
        }
      for (j = 0; j < k; ++j) cc[x][j] += cc[u][j];
      sz[x] += sz[u];
    }
  }
  u = ++cc[x][k - 1];
  for (i = k - 2; i >= 0; --i) cc[x][i + 1] = cc[x][i];
  cc[x][0] = u;
  ret[x] = R + u;
  return ret[x];
}
int main() {
  int n, i, j, u, v, a, b;
  cin >> n >> k;
  for (i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  cout << sum << endl;
}

```