#include <bits/stdc++.h>
using namespace std;
const int maxi = 1e6;
vector<int> v[maxi];
long long sz[maxi];
long long dp[maxi][7], f[100000], sum[10000];
long long tot, ans, x, y, n, k;
void dfs(int x, int pred) {
  sz[x] = 1;
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      dfs(v[x][i], x);
      tot += sz[v[x][i]] * (n - sz[v[x][i]]);
      sz[x] += sz[v[x][i]];
    }
}
void dfs1(int x, int pred) {
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      dfs1(v[x][i], x);
      for (int j = 0; j < k; j++) dp[x][(j + 1) % k] += dp[v[x][i]][j];
    }
  for (int i = 0; i < k; i++) sum[i] = 0;
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      for (int j = 0; j < k; j++)
        for (int l = 0; l < k; l++)
          sum[(l + j + 2) % k] +=
              dp[v[x][i]][j] * (dp[x][(l + 1) % k] - dp[v[x][i]][l]);
    }
  for (int i = 0; i < k; i++) f[i] += sum[i] / 2;
  for (int j = 0; j < k; j++) f[j] += dp[x][j];
  dp[x][0]++;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int i = 1; i < k; i++) ans += f[i];
  for (long long i = 0; i < k; i++) tot -= i * f[i];
  ans += tot / k;
  printf("%I64d\n", ans);
  return 0;
}

### 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 maxi = 1e6;
vector<int> v[maxi];
long long sz[maxi];
long long dp[maxi][7], f[100000], sum[10000];
long long tot, ans, x, y, n, k;
void dfs(int x, int pred) {
  sz[x] = 1;
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      dfs(v[x][i], x);
      tot += sz[v[x][i]] * (n - sz[v[x][i]]);
      sz[x] += sz[v[x][i]];
    }
}
void dfs1(int x, int pred) {
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      dfs1(v[x][i], x);
      for (int j = 0; j < k; j++) dp[x][(j + 1) % k] += dp[v[x][i]][j];
    }
  for (int i = 0; i < k; i++) sum[i] = 0;
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pred) {
      for (int j = 0; j < k; j++)
        for (int l = 0; l < k; l++)
          sum[(l + j + 2) % k] +=
              dp[v[x][i]][j] * (dp[x][(l + 1) % k] - dp[v[x][i]][l]);
    }
  for (int i = 0; i < k; i++) f[i] += sum[i] / 2;
  for (int j = 0; j < k; j++) f[j] += dp[x][j];
  dp[x][0]++;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int i = 1; i < k; i++) ans += f[i];
  for (long long i = 0; i < k; i++) tot -= i * f[i];
  ans += tot / k;
  printf("%I64d\n", ans);
  return 0;
}

```