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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<vector<int>> graph(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  vector<int> seq;
  vector<bool> added(n);
  vector<int> parent(n);
  seq.push_back(0);
  added[0] = true;
  parent[0] = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < graph[seq[i]].size(); j++) {
      if (!added[graph[seq[i]][j]]) {
        parent[graph[seq[i]][j]] = seq[i];
        added[graph[seq[i]][j]] = true;
        seq.push_back(graph[seq[i]][j]);
      }
    }
  }
  reverse(seq.begin(), seq.end());
  vector<long long> dist_sum(n);
  vector<vector<int>> vert_num(n, vector<int>(k));
  vector<long long> subtree_len(n);
  for (int i = 0; i < n; i++) {
    int x = seq[i];
    subtree_len[x] = 1;
    vert_num[x][0] = 1;
    dist_sum[x] = 0;
    for (int j = 0; j < graph[x].size(); j++) {
      if (graph[x][j] != parent[x]) {
        int y = graph[x][j];
        subtree_len[x] += subtree_len[y];
        for (int l = 0; l < k; l++) {
          vert_num[x][(l + 1) % k] += vert_num[y][l];
        }
        dist_sum[x] += dist_sum[y] + subtree_len[y];
      }
    }
  }
  vector<long long> rev_dist_sum(n);
  vector<vector<int>> rev_vert_num(n, vector<int>(k));
  for (int i = n - 1; i >= 0; i--) {
    int x = seq[i];
    if (parent[x] != -1) {
      int y = parent[x];
      for (int l = 0; l < k; l++) {
        rev_vert_num[x][(l + 1) % k] +=
            rev_vert_num[y][l] + vert_num[y][l] - vert_num[x][(l + k - 1) % k];
      }
      rev_dist_sum[x] = rev_dist_sum[y] + n - subtree_len[y] + dist_sum[y] +
                        subtree_len[y] - 2 * subtree_len[x] - dist_sum[x];
    }
  }
  long long answer = 0;
  for (int i = 0; i < n; i++) {
    dist_sum[i] += rev_dist_sum[i];
    for (int j = 0; j < k; j++) {
      dist_sum[i] += (vert_num[i][j] + rev_vert_num[i][j]) * ((k - j) % k);
    }
    answer += dist_sum[i] / k;
  }
  cout << answer / 2 << endl;
  return 0;
}
