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 read() {
  int x = 0, sgn = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') sgn = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
  return x * sgn;
}
const int N = 2e5 + 10;
int n, k, h[N], cnt;
long long ans;
struct edge {
  int v, nxt;
} e[N << 1];
void link(int x, int y) {
  e[++cnt] = (edge){y, h[x]};
  h[x] = cnt;
}
struct stu {
  long long f[6], g[6];
  stu() {
    memset(f, 0, sizeof f);
    memset(g, 0, sizeof g);
  }
  stu operator<<(int x) {
    stu c;
    if (k == 1) {
      c.f[0] = f[0] + g[0];
      c.g[0] = g[0];
      return c;
    }
    for (int i = k - 1; i >= 1; i--) c.f[i] = f[i - 1], c.g[i] = g[i - 1];
    c.f[0] = f[k - 1], c.g[0] = g[k - 1];
    c.f[1] += g[0];
    return c;
  }
  friend stu operator+(stu a, stu b) {
    stu c;
    for (int i = 0; i < k; i++)
      c.f[i] = a.f[i] + b.f[i], c.g[i] = a.g[i] + b.g[i];
    return c;
  }
  friend stu operator-(stu a, stu b) {
    stu c;
    for (int i = 0; i < k; i++)
      c.f[i] = a.f[i] - b.f[i], c.g[i] = a.g[i] - b.g[i];
    return c;
  }
  long long cal() {
    long long res = 0;
    for (int i = 0; i < k; i++) res += f[i];
    return res;
  }
} f[N];
void dfs(int x, int p) {
  f[x].g[0] = 1;
  for (int i = h[x], y; i; i = e[i].nxt)
    if ((y = e[i].v) ^ p) {
      dfs(y, x);
      f[x] = f[x] + (f[y] << 1);
    }
}
void dp(int x, int p) {
  for (int i = h[x], y; i; i = e[i].nxt)
    if ((y = e[i].v) ^ p) {
      f[y] = ((f[x] - (f[y] << 1)) << 1) + f[y];
      dp(y, x);
    }
}
int main() {
  n = read(), k = read();
  for (int i = 1, u, v; i < n; i++)
    u = read(), v = read(), link(u, v), link(v, u);
  dfs(1, 0);
  dp(1, 0);
  for (int i = 1; i <= n; i++) ans += f[i].cal();
  printf("%lld\n", ans >> 1);
  return 0;
}
