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 N = 2e5 + 2;
int n, k, childs[N];
pair<int, int> edges[N];
vector<vector<int> > g(N);
long long a[N][6], ans;
int countChilds(int i, int p) {
  childs[i] = 1;
  for (int j : g[i])
    if (j != p) childs[i] += countChilds(j, i);
  return childs[i];
}
void dfs(int i, int p, int d) {
  ++a[i][d % k];
  for (int j : g[i])
    if (j != p) {
      dfs(j, i, d + 1);
      for (int x = (0); x <= (int)(k - 1); ++x)
        for (int y = (0); y <= (int)(k - 1); ++y) {
          int z = ((x + y - 2 * d) % k + k) % k;
          if (z) ans += a[i][x] * a[j][y] * (k - z);
        }
      for (int x = (0); x <= (int)(k - 1); ++x) a[i][x] += a[j][x];
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (int)(n - 1); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
    edges[i] = make_pair(x, y);
  }
  countChilds(1, 1);
  for (int i = (1); i <= (int)(n - 1); ++i) {
    int mn = min(childs[edges[i].first], childs[edges[i].second]);
    ans += 1ll * mn * (n - mn);
  }
  dfs(1, 1, 0);
  cout << ans / k;
}
