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.
#include <bits/stdc++.h>
const long long maxk = 8;
const long long maxn = 2e5 + 10;
long long k;
long long ans[maxn][maxk];
long long size[maxn][maxk];
long long beitrag[maxn][maxk];
long long bsize[maxn][maxk];
long long cost(long long len) { return (len + k - 1) / k - 2; }
std::vector<long long> v[maxn];
long long tlen = 0;
long long special = 0;
void dfs(long long x, long long p = -1) {
  for (auto next : v[x]) {
    if (next == p) continue;
    dfs(next, x);
    for (long long i = 0; i < k; i++) {
      long long prev = (i - 1 + k) % k;
      beitrag[next][i] = ans[next][prev];
      bsize[next][i] = size[next][prev];
      if (i == 1 || k == 1) beitrag[next][i] += size[next][prev];
      special += beitrag[next][i];
      size[x][i] += bsize[next][i];
      ans[x][i] += beitrag[next][i];
    }
    if (k == 1) {
      beitrag[next][0]++;
      bsize[next][0]++;
      size[x][0]++;
      ans[x][0]++;
    } else {
      beitrag[next][1]++;
      bsize[next][1]++;
      size[x][1]++;
      ans[x][1]++;
    }
    special += 1;
  }
  for (auto next : v[x]) {
    if (next == p) continue;
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        long long tmpj = ans[x][j] - beitrag[next][j];
        long long tmpsize = (size[x][j] - bsize[next][j]);
        if (tmpsize <= 0 || bsize[next][i] <= 0) continue;
        long long merge = cost(i + j);
        if (merge == -1 && ((i != 0 && j != 0) || (i == 0 && j == 0)))
          tlen += (beitrag[next][i] - bsize[next][i]) * tmpsize +
                  tmpj * bsize[next][i];
        else
          tlen += (beitrag[next][i]) * tmpsize + tmpj * bsize[next][i];
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  std::cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    std::cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1);
  std::cout << special + tlen / 2 << std::endl;
  return 0;
}
