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 MAXN = 2e5 + 5;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
int n, k;
int nedge, head[MAXN];
struct Edge {
  int to, nxt;
} edge[MAXN << 1];
int sum[10], f[MAXN][5], siz[MAXN];
long long ans, s[MAXN];
void add(int x, int y) {
  edge[++nedge].to = y;
  edge[nedge].nxt = head[x];
  head[x] = nedge;
}
void dfs(int x, int fa) {
  f[x][0] = 1;
  siz[x] = 1;
  for (int i = head[x]; i; i = edge[i].nxt) {
    int y = edge[i].to;
    if (y == fa) continue;
    dfs(y, x);
    ans += s[x] * siz[y] + s[y] * siz[x];
    for (int j = k - 1; ~j; j--) sum[j] = sum[j + 1] + f[y][j];
    for (int j = 0; j < k; j++) ans += 1LL * f[x][j] * (sum[k - j] + sum[0]);
    for (int j = 0; j < k - 1; j++) f[x][j + 1] += f[y][j];
    s[x] += f[y][k - 1] + s[y];
    f[x][0] += f[y][k - 1];
    siz[x] += siz[y];
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  printf("%lld", ans);
}
