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 maxinput = 200005;
const long long mod = 1000000007;
long long n, k;
vector<long long> adj[maxinput];
long long in[maxinput][5];
long long out[maxinput][5];
long long subtree[maxinput];
long long sum = 0;
void dfs1(long long u, long long par) {
  long long v;
  for (long long i = 0; i < k; i++) in[u][i] = 0;
  in[u][0] = 1;
  subtree[u] = 1;
  for (long long j = 0; j < adj[u].size(); j++) {
    v = adj[u][j];
    if (v != par) {
      dfs1(v, u);
      subtree[u] += subtree[v];
      for (long long i = 0; i < k; i++) in[u][(i + 1) % k] += in[v][i];
    }
  }
  sum += subtree[u] * (n - subtree[u]);
}
void dfs2(long long u, long long par) {
  long long v;
  for (long long i = 0; i < k; i++) out[u][i] = 0;
  if (u != 1) {
    for (long long i = 0; i < k; i++)
      out[u][(i + 1) % k] += out[par][i] + in[par][i] - in[u][(i - 1 + k) % k];
  }
  for (long long i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (v != par) dfs2(v, u);
  }
}
int main() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long residue = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < k; j++) {
      residue += ((k - j) % k) * (in[i][j] + out[i][j]);
    }
  }
  cout << (sum + (residue) / 2) / k;
}
