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>
using namespace std;
long long cnt_up[200100][6], cnt_children[200100][6], ans[200100];
int n, p;
vector<int> g[200100];
void dfs(int k, int prev) {
  for (int i = 0; i < g[k].size(); i++) {
    int to = g[k][i];
    if (to == prev) continue;
    dfs(to, k);
    long long jumps;
    for (int j = 1; j <= p; j++)
      for (int l = 1; l <= p; l++) {
        jumps = (long long)((j + l + p - 1) / p);
        ans[k] += cnt_up[k][j] * cnt_children[to][l - 1] +
                  cnt_children[k][j] * cnt_up[to][l - 1] +
                  jumps * cnt_children[k][j] * cnt_children[to][l - 1];
      }
    for (int j = 1; j <= p; j++)
      ans[k] += cnt_up[to][j - 1] + cnt_children[to][j - 1];
    for (int j = 1; j <= p; j++) cnt_up[k][j] += cnt_up[to][j - 1];
    for (int j = 1; j <= p; j++) cnt_children[k][j] += cnt_children[to][j - 1];
  }
  cnt_up[k][0] = cnt_up[k][p] + cnt_children[k][p];
  cnt_children[k][0] = cnt_children[k][p] + 1ll;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> p;
  int to, from;
  for (int i = 0; i < n - 1; i++) {
    cin >> to >> from;
    g[to].push_back(from);
    g[from].push_back(to);
  }
  dfs(1, 0);
  long long res = 0ll;
  for (int i = 1; i <= n; i++) res += ans[i];
  cout << res << endl;
  return 0;
}
