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 int N = 200005;
vector<int> v[N];
vector<int> depth(N, 0);
int k;
long long res = 0;
int f(int a) {
  if (a >= 0)
    return (a + k - 1) / k;
  else
    return (a / k);
}
vector<array<long long, 2>> dfs(int a, int p = 0) {
  depth[a] = depth[p] + 1;
  vector<array<long long, 2>> x(k, {0, 0});
  x[depth[a] % k] = {depth[a] / k, 1};
  for (int to : v[a]) {
    if (to == p) continue;
    vector<array<long long, 2>> y = dfs(to, a);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        long long now = (x[i][0] * y[j][1] + y[j][0] * x[i][1] -
                         2 * (depth[a] / k) * x[i][1] * y[j][1]);
        int d = i + j - 2 * (depth[a] % k);
        now += (f(d) * x[i][1] * y[j][1]);
        res += now;
      }
    }
    for (int j = 0; j < k; j++) {
      x[j][0] += y[j][0];
      x[j][1] += y[j][1];
    }
  }
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1);
  cout << res << "\n";
  return 0;
}
