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;
const long long inf = 1e16 + 9;
const double PI = 3.14159265;
const long long M = 200005;
const long long MAX = 25;
vector<long long> adj[M];
long long n, k, ans, total_subtree[M], count_subtree[M][5], d[M];
long long subtrac(long long a, long long b) {
  return ((a % k) - (b % k) + k) % k;
}
void dfs(long long x, long long par) {
  count_subtree[x][d[x] % k] = total_subtree[x] = 1;
  for (auto i : adj[x]) {
    if (i != par) {
      d[i] = d[x] + 1;
      dfs(i, x);
      total_subtree[x] += total_subtree[i];
      for (long long q = 0; q < k; q++) {
        for (long long w = 0; w < k; w++) {
          long long remainde = subtrac(q + w, 2 * d[x]);
          long long need = subtrac(k, remainde);
          ans += (need * count_subtree[x][q] * count_subtree[i][w]);
        }
      }
      for (long long j = 0; j < k; j++)
        count_subtree[x][j] += count_subtree[i][j];
    }
  }
  ans += total_subtree[x] * (n - total_subtree[x]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans / k;
}
