#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
int n, k, child[N], pathLen[N][5];
long long sum;
vector<int> adjL[N];
void DFS(int node, int per) {
  for (int ch : adjL[node])
    if (ch != per) {
      DFS(ch, node);
      child[node] += child[ch] + 1;
      sum += 1ll * (n - child[ch] - 1) * (child[ch] + 1);
      for (int i = (0); i <= (int)(k - 1); ++i)
        for (int j = (0); j <= (int)(k - 1); ++j)
          sum += 1ll * (k - (i + j + 1) % k) % k * pathLen[node][i] *
                 pathLen[ch][j];
      for (int i = (0); i <= (int)(k - 1); ++i)
        pathLen[node][(i + 1) % k] += pathLen[ch][i];
    }
  ++pathLen[node][0];
  for (int i = (1); i <= (int)(k - 1); ++i)
    sum += 1ll * (k - i) * pathLen[node][i];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (2); i <= (int)(n); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    adjL[a].push_back(b), adjL[b].push_back(a);
  }
  DFS(1, -1);
  printf("%lld\n", sum / k);
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 = 200001;
int n, k, child[N], pathLen[N][5];
long long sum;
vector<int> adjL[N];
void DFS(int node, int per) {
  for (int ch : adjL[node])
    if (ch != per) {
      DFS(ch, node);
      child[node] += child[ch] + 1;
      sum += 1ll * (n - child[ch] - 1) * (child[ch] + 1);
      for (int i = (0); i <= (int)(k - 1); ++i)
        for (int j = (0); j <= (int)(k - 1); ++j)
          sum += 1ll * (k - (i + j + 1) % k) % k * pathLen[node][i] *
                 pathLen[ch][j];
      for (int i = (0); i <= (int)(k - 1); ++i)
        pathLen[node][(i + 1) % k] += pathLen[ch][i];
    }
  ++pathLen[node][0];
  for (int i = (1); i <= (int)(k - 1); ++i)
    sum += 1ll * (k - i) * pathLen[node][i];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (2); i <= (int)(n); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    adjL[a].push_back(b), adjL[b].push_back(a);
  }
  DFS(1, -1);
  printf("%lld\n", sum / k);
  return 0;
}

```