#include <bits/stdc++.h>
using namespace std;
long long all;
int k;
vector<int> T[200001];
vector<long long> f(int v, int p, vector<long long> A) {
  A[k] += A[0];
  all += A[k];
  int t = A[k - 1];
  for (int i = k - 1; i > 0; --i) A[i] = A[i - 1];
  A[0] = t;
  ++A[0];
  vector<long long> BB(k + 1);
  for (auto i : T[v]) {
    if (i != p) {
      vector<long long> B = f(i, v, A);
      B[k] += B[0];
      t = B[k - 1];
      for (int j = k - 1; j > 0; --j) B[j] = B[j - 1];
      B[0] = t;
      for (int j = 0; j <= k; ++j) {
        A[j] += B[j];
        BB[j] += B[j];
      }
    }
  }
  ++BB[0];
  return BB;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    T[a].push_back(b);
    T[b].push_back(a);
  }
  for (int i = 1;; ++i) {
    if (T[i].size() == 1) {
      f(i, 0, vector<long long>(k + 1));
      break;
    }
  }
  cout << all << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long all;
int k;
vector<int> T[200001];
vector<long long> f(int v, int p, vector<long long> A) {
  A[k] += A[0];
  all += A[k];
  int t = A[k - 1];
  for (int i = k - 1; i > 0; --i) A[i] = A[i - 1];
  A[0] = t;
  ++A[0];
  vector<long long> BB(k + 1);
  for (auto i : T[v]) {
    if (i != p) {
      vector<long long> B = f(i, v, A);
      B[k] += B[0];
      t = B[k - 1];
      for (int j = k - 1; j > 0; --j) B[j] = B[j - 1];
      B[0] = t;
      for (int j = 0; j <= k; ++j) {
        A[j] += B[j];
        BB[j] += B[j];
      }
    }
  }
  ++BB[0];
  return BB;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    T[a].push_back(b);
    T[b].push_back(a);
  }
  for (int i = 1;; ++i) {
    if (T[i].size() == 1) {
      f(i, 0, vector<long long>(k + 1));
      break;
    }
  }
  cout << all << endl;
  return 0;
}

```