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;
int n, k, tot, ls[300000];
long long ans, dp[300000][5], h[300000];
struct arr {
  int x, y, next;
} edge[300000 * 2];
void add(int x, int y) {
  edge[++tot].x = x;
  edge[tot].y = y;
  edge[tot].next = ls[x];
  ls[x] = tot;
}
void dfs(int x, int fa, int d) {
  h[x] = dp[x][d] = 1;
  for (int i = ls[x]; i; i = edge[i].next) {
    int y = edge[i].y;
    if (y == fa) continue;
    dfs(y, x, (d + 1) % k);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        int l = (k - (i + j - d * 2) % k) % k;
        ans += l * dp[x][i] * dp[y][j];
      }
    }
    for (int i = 0; i < k; i++) dp[x][i] += dp[y][i];
    ans += (n - h[y]) * h[y];
    h[x] += h[y];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, 0);
  printf("%lld\n", ans / k);
}
