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;
vector<int> g[200111];
long long dp[200111][5];
long long subsz[200111][5];
long long ans = 0ll;
int n, k;
void dfs(int v, int pre) {
  int i, j;
  for (auto nextv : g[v]) {
    if (nextv != pre) {
      vector<int> tmpsz = vector<int>(k, 0);
      vector<long long> tmpcnt = vector<long long>(k, 0ll);
      dfs(nextv, v);
      for (i = 0; i < k; i++) {
        tmpsz[i] += subsz[nextv][((i - 1) + k) % k];
        if (i == 1 || k == 1) {
          tmpcnt[i] +=
              dp[nextv][((i - 1) + k) % k] + subsz[nextv][((i - 1) + k) % k];
        } else {
          tmpcnt[i] += dp[nextv][((i - 1) + k) % k];
        }
      }
      for (i = 0; i < k; i++) {
        for (j = 0; j < k; j++) {
          ans += dp[v][i] * tmpsz[j] + tmpcnt[j] * subsz[v][i];
          if (i > 0 && j > 0 && i + j <= k) {
            ans = ans - subsz[v][i] * tmpsz[j];
          }
        }
      }
      for (i = 0; i < k; i++) {
        dp[v][i] += tmpcnt[i];
        subsz[v][i] += tmpsz[i];
      }
    }
  }
  for (i = 0; i < k; i++) {
    ans = ans + dp[v][i];
  }
  subsz[v][0]++;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int v1, v2;
    scanf("%d%d", &v1, &v2);
    g[v1].push_back(v2);
    g[v2].push_back(v1);
  }
  dfs(1, -1);
  printf("%lld\n", ans);
  return 0;
}
