#include <bits/stdc++.h>
using namespace std;
long long cnt[200100][5], tot[200100], ans = 0;
int k, n;
vector<int> g[200100];
void dfs(int u, int p, int d) {
  int dmodk, res;
  tot[u] = cnt[u][d % k] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    if (g[u][i] != p) {
      dfs(g[u][i], u, d + 1);
      for (int j = 0; j < k; j++) {
        for (int l = 0; l < k; l++) {
          dmodk = ((j + l - 2 * d) % k + k) % k;
          res = (((k - dmodk) % k) + k) % k;
          ans += 1ll * res * cnt[u][j] * cnt[g[u][i]][l];
        }
      }
      for (int j = 0; j < k; j++) cnt[u][j] += cnt[g[u][i]][j];
      tot[u] += tot[g[u][i]];
    }
  }
  ans += tot[u] * (n - tot[u]);
}
int main() {
  int i, u, v;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0);
  printf("%I64d\n", ans / k);
}

### 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;
long long cnt[200100][5], tot[200100], ans = 0;
int k, n;
vector<int> g[200100];
void dfs(int u, int p, int d) {
  int dmodk, res;
  tot[u] = cnt[u][d % k] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    if (g[u][i] != p) {
      dfs(g[u][i], u, d + 1);
      for (int j = 0; j < k; j++) {
        for (int l = 0; l < k; l++) {
          dmodk = ((j + l - 2 * d) % k + k) % k;
          res = (((k - dmodk) % k) + k) % k;
          ans += 1ll * res * cnt[u][j] * cnt[g[u][i]][l];
        }
      }
      for (int j = 0; j < k; j++) cnt[u][j] += cnt[g[u][i]][j];
      tot[u] += tot[g[u][i]];
    }
  }
  ans += tot[u] * (n - tot[u]);
}
int main() {
  int i, u, v;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0);
  printf("%I64d\n", ans / k);
}

```