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;
struct Node {
  Node() : sum(0), steps(0) {
    mod.resize(5, 0);
    mod[0] = 1;
  }
  vector<uint64_t> adj;
  vector<uint64_t> mod;
  uint64_t steps;
  uint64_t sum;
};
void dfs(vector<Node>& a, int curr, int p, int k) {
  for (auto e : a[curr].adj) {
    if (e != p) {
      dfs(a, e, curr, k);
      int tmp = a[e].mod[k - 1];
      for (int i = k - 1; i >= 1; i--) {
        a[e].mod[i] = a[e].mod[i - 1];
      }
      a[e].steps += tmp;
      a[e].mod[0] = tmp;
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          if (i + j > k)
            a[curr].sum += 2 * a[curr].mod[i] * a[e].mod[j];
          else if (i + j > 0)
            a[curr].sum += a[curr].mod[i] * a[e].mod[j];
        }
      }
      for (int i = 0; i < k; i++) {
        a[curr].sum += a[curr].mod[i] * (a[e].steps);
      }
      for (int j = 0; j < k; j++) {
        a[curr].sum += a[e].mod[j] * (a[curr].steps);
      }
      a[curr].sum += a[e].sum;
      a[curr].steps += a[e].steps;
      for (int i = 0; i < k; i++) {
        a[curr].mod[i] += a[e].mod[i];
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<Node> a(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    a[u - 1].adj.push_back(v - 1);
    a[v - 1].adj.push_back(u - 1);
  }
  dfs(a, 0, -1, k);
  cout << a[0].sum << endl;
  return 0;
}
