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 = 200 * 1000 + 10;
const int K = 6;
int n, k;
vector<int> g[N];
long long res = 0;
struct Data {
  long long sum_at[K];
  long long cnt_at[K];
  Data() {
    fill(sum_at, sum_at + K, 0);
    fill(cnt_at, cnt_at + K, 0);
  }
  Data shift() const {
    Data new_data;
    for (int i = 0; i < k; i++) {
      new_data.sum_at[i] = sum_at[(i - 1 + k) % k];
      new_data.cnt_at[i] = cnt_at[(i - 1 + k) % k];
    }
    if (k > 1)
      new_data.sum_at[1] += new_data.cnt_at[1];
    else
      new_data.sum_at[0] += new_data.cnt_at[0];
    return new_data;
  }
  void unite(Data rhs) {
    rhs = rhs.shift();
    for (int m1 = 0; m1 < k; m1++)
      for (int m2 = 0; m2 < k; m2++) {
        res += sum_at[m1] * rhs.cnt_at[m2] + cnt_at[m1] * rhs.sum_at[m2];
        if (k > 1 && m1 && m2 && m1 + m2 <= k)
          res -= cnt_at[m1] * rhs.cnt_at[m2];
      }
    for (int i = 0; i < k; i++) {
      sum_at[i] += rhs.sum_at[i];
      cnt_at[i] += rhs.cnt_at[i];
    }
  }
  void print() {
    for (int i = 0; i < k; i++) cerr << cnt_at[i] << " " << sum_at[i] << endl;
    cerr << endl;
  }
};
Data dfs(int v, int par) {
  Data res;
  res.cnt_at[0]++;
  for (int to : g[v])
    if (to != par) {
      res.unite(dfs(to, v));
    }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  cout << res << endl;
}
