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 count_subtree[200005][5], tot_subtree[200005];
vector<int> adj[200005];
int n, k;
long long ans = 0;
inline void dfs(int c, int p, int d) {
  count_subtree[c][d % k] = tot_subtree[c] = 1;
  for (auto it : adj[c]) {
    if (it != p) {
      dfs(it, c, d + 1);
      for (int x = 0; x < k; x++)
        for (int y = 0; y < k; y++) {
          int rem = ((x + y) - (2 * d) + k) % k;
          int actual = (k - rem + k) % k;
          ans += 1LL * actual * count_subtree[c][x] * count_subtree[it][y];
        }
      for (int x = 0; x < k; x++) count_subtree[c][x] += count_subtree[it][x];
      tot_subtree[c] += tot_subtree[it];
    }
  }
  ans += 1LL * tot_subtree[c] * (n - tot_subtree[c]);
}
int main() {
  scanf("%d", &n);
  ;
  scanf("%d", &k);
  ;
  for (int x = 0; x < n - 1; x++) {
    int u, v;
    scanf("%d", &u);
    ;
    scanf("%d", &v);
    ;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  dfs(1, 0, 0);
  ans /= k;
  printf("%lld", ans);
  ;
}
