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;
mt19937 gen(time(NULL));
const double eps = 1e-9;
const int inf = 1e9;
const int mod = 1e9 + 7;
long long d[200050][5];
long long p2[200050][5];
vector<int> g[200050];
long long ans = 0;
int n, k;
int s[200050];
int dfs(int u, int e) {
  s[u] = 0;
  for (auto v : g[u])
    if (v != e) s[u] += dfs(v, u);
  for (auto v : g[u])
    if (v != e)
      for (int j = 0; j < k; j++) p2[u][j] += p2[v][j], d[u][j] += d[v][j];
  long long la1 = 0, la2 = 0;
  for (auto v : g[u])
    if (v != e)
      for (int j = 0; j < k; j++) {
        la2 += (s[u] - s[v]) * d[v][j];
        for (int q = 0; q < k; q++) {
          if (!q && !j) continue;
          if (q + j <= k)
            la1 += p2[v][j] * (p2[u][q] - p2[v][q]);
          else
            la1 += 2 * p2[v][j] * (p2[u][q] - p2[v][q]);
        }
      }
  la1 /= 2;
  ans += la1 + la2;
  for (auto v : g[u])
    for (int j = 0; j < k; j++) ans += d[v][j] + (j ? p2[v][j] : 0);
  for (int j = 0; j < k; j++) p2[u][j] = 0, d[u][j] = 0;
  if (k > 1)
    p2[u][1]++;
  else
    p2[u][0]++, d[u][0]++;
  for (auto v : g[u]) {
    for (int j = 0; j < k; j++)
      if (j) {
        d[u][j] += d[v][j - 1];
        p2[u][j] += p2[v][j - 1];
      }
    p2[u][0] += p2[v][k - 1];
    d[u][0] += d[v][k - 1];
    d[u][0] += p2[v][k - 1];
  }
  return ++s[u];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
  }
  dfs(1, 0);
  cout << ans << '\n';
}
