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;
void init() {}
const int N = (int)2e5 + 34;
vector<int> es[N];
int n, k;
long long num[N][6];
long long ful[N][6];
long long ans = 0;
void go(int v, int pr) {
  if (es[v].size() == 0) {
    return;
  }
  for (int i = 0; i < (int)es[v].size(); i++)
    if (es[v][i] == pr) swap(es[v][i], es[v].back());
  if (es[v].back() == pr) es[v].pop_back();
  for (auto w : es[v]) {
    go(w, v);
    for (int i = k - 1; i >= 0; i--) {
      num[w][i + 1] = num[w][i];
      ful[w][i + 1] = ful[w][i];
    }
    num[w][1]++;
    num[w][0] = num[w][k];
    ful[w][0] = ful[w][k] + num[w][k];
    for (int i = 0; i < k; i++) {
      num[v][i] += num[w][i];
      ful[v][i] += ful[w][i];
    }
  }
  for (int o = 0; o < k; o++) {
    ans += 2 * (ful[v][o] + ((o + k - 1) / k) * num[v][o]);
  }
  long long num1, ful1;
  for (auto w : es[v]) {
    for (int o0 = 0; o0 < k; o0++) {
      for (int o1 = 0; o1 < k; o1++) {
        num1 = num[v][o1] - num[w][o1];
        ful1 = ful[v][o1] - ful[w][o1];
        ans += ful[w][o0] * num1 + ful1 * num[w][o0];
        ans += ((o0 + o1 + k - 1) / k) * num1 * num[w][o0];
      }
    }
  }
}
int main() {
  init();
  int v1, v2;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &v1, &v2);
    v1--;
    v2--;
    es[v1].push_back(v2);
    es[v2].push_back(v1);
  }
  go(0, -1);
  printf("%I64d\n", ans / 2);
  return 0;
}
