#include <bits/stdc++.h>
using namespace std;
long long n, d, fq[200069][6], dp[200069][6], z = 0;
vector<long long> al[200069];
bitset<200069> vtd;
void dfs(long long x) {
  long long i, j, r, sz = al[x].size(), l, tf[6], td[6];
  vtd[x] = 1;
  fq[x][d] = 1;
  dp[x][d] = -1;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      dfs(l);
      for (j = 1; j <= d; j++) {
        tf[j] = 0;
        td[j] = 0;
      }
      for (j = 1; j <= d; j++) {
        tf[j % d + 1] += fq[l][j];
        td[j % d + 1] += dp[l][j] + fq[l][j] * (j / d);
      }
      for (j = 1; j <= d; j++) {
        fq[l][j] = tf[j];
        dp[l][j] = td[j];
      }
      for (j = 1; j <= d; j++) {
        for (r = 1; r <= d; r++) {
          z += dp[x][j] * fq[l][r] + dp[l][r] * fq[x][j] +
               fq[x][j] * fq[l][r] * ((j + r - 1) / d + 1);
        }
      }
      for (j = 1; j <= d; j++) {
        fq[x][j] += fq[l][j];
        dp[x][j] += dp[l][j];
      }
    }
  }
}
int main() {
  long long i, k, l;
  scanf("%lld%lld", &n, &d);
  for (i = 0; i < n - 1; i++) {
    scanf("%lld%lld", &k, &l);
    al[k].push_back(l);
    al[l].push_back(k);
  }
  dfs(1);
  printf("%lld", z);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, d, fq[200069][6], dp[200069][6], z = 0;
vector<long long> al[200069];
bitset<200069> vtd;
void dfs(long long x) {
  long long i, j, r, sz = al[x].size(), l, tf[6], td[6];
  vtd[x] = 1;
  fq[x][d] = 1;
  dp[x][d] = -1;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      dfs(l);
      for (j = 1; j <= d; j++) {
        tf[j] = 0;
        td[j] = 0;
      }
      for (j = 1; j <= d; j++) {
        tf[j % d + 1] += fq[l][j];
        td[j % d + 1] += dp[l][j] + fq[l][j] * (j / d);
      }
      for (j = 1; j <= d; j++) {
        fq[l][j] = tf[j];
        dp[l][j] = td[j];
      }
      for (j = 1; j <= d; j++) {
        for (r = 1; r <= d; r++) {
          z += dp[x][j] * fq[l][r] + dp[l][r] * fq[x][j] +
               fq[x][j] * fq[l][r] * ((j + r - 1) / d + 1);
        }
      }
      for (j = 1; j <= d; j++) {
        fq[x][j] += fq[l][j];
        dp[x][j] += dp[l][j];
      }
    }
  }
}
int main() {
  long long i, k, l;
  scanf("%lld%lld", &n, &d);
  for (i = 0; i < n - 1; i++) {
    scanf("%lld%lld", &k, &l);
    al[k].push_back(l);
    al[l].push_back(k);
  }
  dfs(1);
  printf("%lld", z);
}

```