#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
int n, k;
ll res;
const int MAX = 2e5 + 10;
ll s[MAX][5];
ll q[MAX][5];
vi e[MAX];
int v[MAX];
void dfs(int x) {
  v[x] = 1;
  q[x][0] += 1;
  for (int y : e[x]) {
    if (v[y]) continue;
    dfs(y);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        res += s[x][i] * q[y][j];
        res += s[y][j] * q[x][i];
        int c1 = (i + j + 1) / k;
        if ((i + j + 1) % k) c1++;
        int c2 = i / k + j / k;
        if (i % k) c2++;
        if (j % k) c2++;
        res += q[x][i] * q[y][j] * (c1 - c2);
      }
    }
    for (int i = 0; i < k; i++) {
      q[x][(i + 1) % k] += q[y][i];
      s[x][(i + 1) % k] += s[y][i];
    }
    s[x][1 % k] += q[y][0];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dfs(1);
  cout << res << "\n";
}

### Prompt

Create a solution in cpp for 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;
using ll = long long;
using vi = vector<int>;
int n, k;
ll res;
const int MAX = 2e5 + 10;
ll s[MAX][5];
ll q[MAX][5];
vi e[MAX];
int v[MAX];
void dfs(int x) {
  v[x] = 1;
  q[x][0] += 1;
  for (int y : e[x]) {
    if (v[y]) continue;
    dfs(y);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        res += s[x][i] * q[y][j];
        res += s[y][j] * q[x][i];
        int c1 = (i + j + 1) / k;
        if ((i + j + 1) % k) c1++;
        int c2 = i / k + j / k;
        if (i % k) c2++;
        if (j % k) c2++;
        res += q[x][i] * q[y][j] * (c1 - c2);
      }
    }
    for (int i = 0; i < k; i++) {
      q[x][(i + 1) % k] += q[y][i];
      s[x][(i + 1) % k] += s[y][i];
    }
    s[x][1 % k] += q[y][0];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dfs(1);
  cout << res << "\n";
}

```