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;
long long n, k, u, v, dp[200010][5], len[200010][5], child[200010][5],
    num[200010][5], rem[5], ram[5], ans = 0;
vector<long long> graph[200010];
void dfs(long long u) {
  memset(rem, 0, sizeof(rem));
  memset(ram, 0, sizeof(ram));
  long long p = -1, s;
  for (long long i = 0; i < k; i++) child[u][i] = dp[u][i] = len[u][i] = 0;
  for (long long i = 0; i < graph[u].size(); i++) {
    long long v = graph[u][i];
    if (dp[v][0] == -1) {
      dfs(v);
      child[u][0] += child[v][k - 1],
          dp[u][0] += dp[v][k - 1] + child[v][k - 1],
          len[u][0] += len[v][0] + dp[v][0],
          num[u][0] += num[v][0] + child[v][0];
      for (long long j = 1; j < k; j++) {
        child[u][j] += child[v][j - 1],
            dp[u][j] += dp[v][j - 1] + child[v][j - 1],
            len[u][j] += len[v][j] + dp[v][j],
            num[u][j] += num[v][j] + child[v][j];
      }
    } else
      p = v;
  }
  if (p == -1 || graph[u].size() > 2) {
    for (long long j = 0; j < graph[u].size(); j++) {
      long long v = graph[u][j];
      if (v != p) {
        for (long long i = 0; i < k; i++)
          rem[i] = dp[v][i] + child[v][i], ram[i] = child[v][i];
        s = j + 1;
        break;
      }
    }
    for (long long i = s; i < graph[u].size(); i++) {
      long long v = graph[u][i];
      if (v != p) {
        for (long long j = 0; j < k; j++) {
          for (long long l = 0; l < k; l++)
            len[u][(j + l + 2) % k] +=
                rem[j] * child[v][l] + (dp[v][l] + child[v][l]) * ram[j],
                num[u][(j + l + 2) % k] += ram[j] * child[v][l];
        }
        for (long long j = 0; j < k; j++)
          rem[j] += dp[v][j] + child[v][j], ram[j] += child[v][j];
      }
    }
  }
  child[u][0]++;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    scanf("%lld%lld", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(1);
  ans = len[1][0] + dp[1][0];
  for (long long i = 1; i < k; i++)
    ans += len[1][i] + dp[1][i] + (num[1][i] + child[1][i]) * (k - i);
  printf("%lld", ans / k);
}
