#include <bits/stdc++.h>
using namespace std;
int n, k;
long long ans;
vector<vector<int> > adjList;
vector<pair<long long, long long> > dfs(int node, int parent) {
  vector<pair<long long, long long> > ret(k + 1);
  for (int i = 0; i <= k; i++) ret[i].first = 0, ret[i].second = 0;
  for (int ind = 0; ind < adjList[node].size(); ind++) {
    int ch = adjList[node][ind];
    if (ch != parent) {
      vector<pair<long long, long long> > temp = dfs(ch, node);
      pair<long long, long long> t = temp[k];
      for (int i = k - 1; i >= 0; i--)
        temp[i + 1] = temp[i], ans += temp[i].first + temp[i].second;
      temp[1].first += t.first;
      temp[1].second += t.first + t.second;
      ans += 2 * t.first;
      ans += t.second;
      for (int i = 1; i <= k; i++) {
        for (int j = 1; j <= k; j++) {
          ans += temp[i].first * ret[j].first + temp[i].second * ret[j].first +
                 ret[j].second * temp[i].first;
          if (i + j > k) ans += temp[i].first * ret[j].first;
        }
      }
      for (int i = 1; i <= k; i++)
        ret[i].first += temp[i].first, ret[i].second += temp[i].second;
    }
  }
  ret[0].first++;
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  adjList.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    adjList[x].push_back(y);
    adjList[y].push_back(x);
  }
  dfs(1, 0);
  printf("%I64d", ans);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
int n, k;
long long ans;
vector<vector<int> > adjList;
vector<pair<long long, long long> > dfs(int node, int parent) {
  vector<pair<long long, long long> > ret(k + 1);
  for (int i = 0; i <= k; i++) ret[i].first = 0, ret[i].second = 0;
  for (int ind = 0; ind < adjList[node].size(); ind++) {
    int ch = adjList[node][ind];
    if (ch != parent) {
      vector<pair<long long, long long> > temp = dfs(ch, node);
      pair<long long, long long> t = temp[k];
      for (int i = k - 1; i >= 0; i--)
        temp[i + 1] = temp[i], ans += temp[i].first + temp[i].second;
      temp[1].first += t.first;
      temp[1].second += t.first + t.second;
      ans += 2 * t.first;
      ans += t.second;
      for (int i = 1; i <= k; i++) {
        for (int j = 1; j <= k; j++) {
          ans += temp[i].first * ret[j].first + temp[i].second * ret[j].first +
                 ret[j].second * temp[i].first;
          if (i + j > k) ans += temp[i].first * ret[j].first;
        }
      }
      for (int i = 1; i <= k; i++)
        ret[i].first += temp[i].first, ret[i].second += temp[i].second;
    }
  }
  ret[0].first++;
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  adjList.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    adjList[x].push_back(y);
    adjList[y].push_back(x);
  }
  dfs(1, 0);
  printf("%I64d", ans);
  return 0;
}

```