#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> gr[200005];
long long ans = 0;
int child[200005], dist[200005][6];
void go(int now, int par) {
  child[now] = 1;
  int i, j;
  for (auto to : gr[now]) {
    if (to == par) continue;
    go(to, now);
    ans += ((long long)child[to] * (n - child[to]));
    child[now] += child[to];
    for (i = 0; i < k; i++) {
      for (j = 0; j < k; j++) {
        int x = (i + j + 1) % k;
        if (!x) continue;
        ans += ((long long)dist[now][i] * dist[to][j]) * (k - x);
      }
    }
    for (i = 1; i < k; i++) {
      dist[now][i] += dist[to][i - 1];
      ans += ((long long)dist[to][i - 1] * (k - i));
    }
    dist[now][0] += dist[to][k - 1];
  }
  dist[now][0]++;
}
int main() {
  int i, j;
  scanf("%d", &n);
  scanf("%d", &k);
  for (i = 1; i < n; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  go(1, -1);
  cout << ans / k << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int n, k;
vector<int> gr[200005];
long long ans = 0;
int child[200005], dist[200005][6];
void go(int now, int par) {
  child[now] = 1;
  int i, j;
  for (auto to : gr[now]) {
    if (to == par) continue;
    go(to, now);
    ans += ((long long)child[to] * (n - child[to]));
    child[now] += child[to];
    for (i = 0; i < k; i++) {
      for (j = 0; j < k; j++) {
        int x = (i + j + 1) % k;
        if (!x) continue;
        ans += ((long long)dist[now][i] * dist[to][j]) * (k - x);
      }
    }
    for (i = 1; i < k; i++) {
      dist[now][i] += dist[to][i - 1];
      ans += ((long long)dist[to][i - 1] * (k - i));
    }
    dist[now][0] += dist[to][k - 1];
  }
  dist[now][0]++;
}
int main() {
  int i, j;
  scanf("%d", &n);
  scanf("%d", &k);
  for (i = 1; i < n; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  go(1, -1);
  cout << ans / k << endl;
}

```