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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int M = 2e5 + 5, N = 6;
int n, k;
vector<int> g[M];
long long d[M][N];
long long ans[N], dv[M], an, anl[M][N];
void dfs(int v, int pr = -1) {
  d[v][0] = 1;
  dv[v] = 1;
  for (int i = 0; i < g[v].size(); ++i)
    if (g[v][i] != pr) dfs(g[v][i], v), dv[v] += dv[g[v][i]];
  for (int j = 0; j < k; ++j)
    for (int i = 0; i < g[v].size(); ++i)
      if (g[v][i] != pr)
        d[v][(j + 1) % k] += d[g[v][i]][j], ans[(j + 1) % k] += d[g[v][i]][j];
  for (int j = 0; j < k; ++j)
    for (int h = 0; h < k; ++h)
      for (int i = 0; i < g[v].size(); ++i)
        if (g[v][i] != pr) {
          int x = d[g[v][i]][(h - 1 + k) % k];
          if (h == 0) x += 1;
          anl[v][(j + 1 + h) % k] += d[g[v][i]][j] * (d[v][h] - x);
        }
  for (int i = 0; i < k; ++i) ans[i] += anl[v][i] / 2;
  an += dv[v] * (n - dv[v]);
}
int main() {
  cin >> n >> k;
  int v, u;
  for (int i = 0; i < n - 1; ++i) {
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(1);
  for (int j = 1; j < k; ++j) an += ans[j] * ((-j + k) % k);
  cout << an / k;
}
