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 = 2e5 + 5;
int n, k, u, v;
long long dp[N][5];
long long ans = 0;
vector<vector<int> > T;
int dfs(int u, int v) {
  int subt_sz = 1;
  fill(dp[v], dp[v] + k, 0);
  dp[v][0] = 1;
  for (int w : T[v]) {
    if (u == w) continue;
    subt_sz += dfs(v, w);
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        ans += ((2 * k - i - j - 1) % k) * dp[v][i] * dp[w][j];
    for (int i = 0; i < k; ++i) dp[v][(i + 1) % k] += dp[w][i];
  }
  ans += 1LL * subt_sz * (n - subt_sz);
  return subt_sz;
}
int main() {
  scanf("%d%d", &n, &k);
  T = vector<vector<int> >(n, vector<int>());
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    T[--u].push_back(--v);
    T[v].push_back(u);
  }
  dfs(-1, 0);
  printf("%I64d\n", ans / k);
}
