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> adj[N];
long long int n, k;
long long int dp[N][6], cou[N][6], conti[N][6], contic[N][6], tdp[6], tcdp[6],
    ans;
void find_ans(int v, int p) {
  for (auto it : adj[v]) {
    if (it != p) {
      find_ans(it, v);
      for (int j = 1; j <= k; j++) {
        contic[it][j] = cou[it][(j == 1 ? k : j - 1)] + (j == 1);
        conti[it][j] =
            dp[it][(j == 1 ? k : j - 1)] + (j == 1) * (cou[it][k] + 1);
      }
      for (int j = 1; j <= k; j++) {
        dp[v][j] += conti[it][j];
        cou[v][j] += contic[it][j];
      }
    }
  }
  for (int j = 1; j <= k; j++) {
    tdp[j] = dp[v][j];
    tcdp[j] = cou[v][j];
    ans += dp[v][j];
  }
  for (auto it : adj[v]) {
    if (it != p) {
      for (int j = 1; j <= k; j++) {
        tdp[j] -= conti[it][j];
        tcdp[j] -= contic[it][j];
      }
      for (int j = 1; j <= k; j++) {
        for (int j1 = 1; j1 <= k; j1++) {
          long long int c = tcdp[j] * contic[it][j1];
          if (c) {
            ans += max(0LL, tdp[j] - tcdp[j]) * contic[it][j1] +
                   max(0LL, conti[it][j1] - contic[it][j1]) * tcdp[j] +
                   ((j1 + j + k - 1) / k) * c;
          }
        }
      }
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  find_ans(1, 0);
  cout << ans;
}
