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 {
  vector<int> neighs;
};
int k;
vector<Node> nodes;
long long ans = 0;
vector<vector<long long>> s, c;
void DFS(int root, int parent) {
  for (int u : nodes[root].neighs) {
    if (u == parent) continue;
    DFS(u, root);
  }
  for (int mod = 0; mod < k; mod++) {
    int next_mod = (mod + 1) % k;
    for (int i : nodes[root].neighs) {
      if (i == parent) continue;
      s[next_mod][root] += s[mod][i];
      if (mod == k - 1) s[next_mod][root] += c[mod][i];
      c[next_mod][root] += c[mod][i];
    }
  }
  c[0][root]++;
  for (int i = 0; i < k; i++) {
    long long junk = (i + 1) / k;
    if ((i + 1) % k != 0) junk++;
    for (int u : nodes[root].neighs) {
      if (u == parent) continue;
      ans += junk * c[i][u] + s[i][u];
    }
    for (int j = 0; j < k; j++) {
      junk = (i + j + 2) / k;
      if ((i + j + 2) % k != 0) junk++;
      long long part_sum_ci = 0;
      long long part_sum_si = 0;
      for (int u : nodes[root].neighs) {
        if (u == parent) continue;
        ans += junk * c[j][u] * part_sum_ci;
        ans += s[j][u] * part_sum_ci + c[j][u] * part_sum_si;
        part_sum_ci += c[i][u];
        part_sum_si += s[i][u];
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n >> k;
  nodes.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    nodes[a].neighs.push_back(b);
    nodes[b].neighs.push_back(a);
  }
  s.resize(k);
  c.resize(k);
  for (int mod = 0; mod < k; mod++) {
    s[mod].resize(n);
    c[mod].resize(n);
  }
  DFS(0, -1);
  cout << ans << '\n';
  return 0;
}
