#include <bits/stdc++.h>
using namespace std;
long long n, k, x, y, kk, ans, f[500005][10], dp[500005][10], head[500005];
struct Tree {
  long long nxt, to;
} e[500005];
inline void link(long long x, long long y) {
  e[++kk].nxt = head[x];
  e[kk].to = y;
  head[x] = kk;
}
void dfs1(long long u, long long fa) {
  dp[u][0]++;
  for (long long i = head[u]; i; i = e[i].nxt) {
    long long v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    for (long long i1 = 0; i1 <= k; i1++) {
      for (long long j = 0; j <= k; j++) {
        long long tmp = i1 + j + 1;
        if (tmp % k)
          tmp = tmp / k + 1;
        else
          tmp = tmp / k;
        ans = ans + dp[u][i1] * dp[v][j] * tmp;
        ans = ans + dp[u][i1] * f[v][j] + dp[v][j] * f[u][i1];
      }
    }
    for (long long j = 1; j <= k; j++)
      dp[u][j] += dp[v][j - 1], f[u][j] += f[v][j - 1];
    dp[u][1] += dp[v][k];
    f[u][1] += f[v][k] + dp[v][k];
  }
}
signed main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    scanf("%lld%lld", &x, &y);
    link(x, y);
    link(y, x);
  }
  dfs1(1, -1);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, k, x, y, kk, ans, f[500005][10], dp[500005][10], head[500005];
struct Tree {
  long long nxt, to;
} e[500005];
inline void link(long long x, long long y) {
  e[++kk].nxt = head[x];
  e[kk].to = y;
  head[x] = kk;
}
void dfs1(long long u, long long fa) {
  dp[u][0]++;
  for (long long i = head[u]; i; i = e[i].nxt) {
    long long v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    for (long long i1 = 0; i1 <= k; i1++) {
      for (long long j = 0; j <= k; j++) {
        long long tmp = i1 + j + 1;
        if (tmp % k)
          tmp = tmp / k + 1;
        else
          tmp = tmp / k;
        ans = ans + dp[u][i1] * dp[v][j] * tmp;
        ans = ans + dp[u][i1] * f[v][j] + dp[v][j] * f[u][i1];
      }
    }
    for (long long j = 1; j <= k; j++)
      dp[u][j] += dp[v][j - 1], f[u][j] += f[v][j - 1];
    dp[u][1] += dp[v][k];
    f[u][1] += f[v][k] + dp[v][k];
  }
}
signed main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    scanf("%lld%lld", &x, &y);
    link(x, y);
    link(y, x);
  }
  dfs1(1, -1);
  printf("%lld\n", ans);
  return 0;
}

```