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 mod = 1000000007;
int n, k;
long long L;
vector<vector<int>> g;
vector<long long> total;
vector<int> visited;
vector<vector<long long>> cnt;
int dfs(int cur) {
  visited[cur] = 1;
  long long r = 1;
  for (int n : g[cur]) {
    if (visited[n] == 0) {
      r += dfs(n);
      ++cnt[n][0];
      for (int i = 0; i < k; ++i) {
        for (int j = 0; j < k; ++j) {
          total[(i + j + 1) % k] += cnt[cur][i] * cnt[n][j];
        }
      }
      --cnt[n][0];
      for (int i = 0; i < k; ++i) {
        cnt[cur][(i + 1) % k] += cnt[n][i];
      }
      ++cnt[cur][1 % k];
    }
  }
  for (int i = 0; i < k; ++i) total[i] += cnt[cur][i];
  L += r * (n - r);
  return r;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  g.resize(n + 1);
  total.resize(k, 0);
  visited.resize(n + 1, 0);
  cnt = vector<vector<long long>>(n + 1, vector<long long>(k, 0));
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  L = 0;
  dfs(1);
  for (int i = 1; i < k; ++i) L += total[i] * (k - i);
  cout << L / k << endl;
  return 0;
}
