#include <bits/stdc++.h>
using namespace std;
int k;
void solve(int v, int p, vector<vector<long long> > &up,
           vector<vector<long long> > &down, vector<long long> &ans,
           vector<vector<long long> > &graph, vector<long long> &ans_d) {
  if (p != -1) {
    for (int i = 1; i < up[p].size(); ++i) {
      up[v][i] += up[p][i - 1];
    }
    up[v][0] += up[p].back();
    up[v][min(1, k - 1)]++;
    ans[v] += ans[p];
    ans[v] += up[p][0];
    ans[v]++;
    for (int i = 1; i < down[p].size(); ++i) {
      up[v][i] += down[p][i - 1];
    }
    up[v][0] += down[p].back();
    if (down[p][min(k - 1, 1)]) {
      up[v][min(1, k - 1)] + down[p][0];
    }
    ans[v] += ans_d[p];
    ans[v] += down[p][0];
  }
  for (int i = 0; i < graph[v].size(); ++i) {
    long long to = graph[v][i];
    if (to == p) {
      continue;
    }
    solve(to, v, up, down, ans, graph, ans_d);
    ans_d[v] += ans_d[to];
    ans_d[v] += down[to][0];
    ans_d[v] += 1;
    for (int i = 1; i < down[to].size(); ++i) {
      down[v][i] += down[to][i - 1];
    }
    down[v][0] += down[to].back();
    down[v][min(k - 1, 1)]++;
  }
}
int main() {
  long long n;
  cin >> n >> k;
  vector<vector<long long> > graph(n);
  long long from, to;
  for (int i = 0; i < n - 1; ++i) {
    cin >> from >> to;
    graph[from - 1].push_back(to - 1);
    graph[to - 1].push_back(from - 1);
  }
  vector<vector<long long> > up(n, vector<long long>(k)),
      down(n, vector<long long>(k));
  vector<long long> ans(n), ans_d(n);
  solve(0, -1, up, down, ans, graph, ans_d);
  long long result = 0;
  for (int i = 0; i < n; ++i) {
    result += ans[i];
  }
  cout << result;
}

### 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;
int k;
void solve(int v, int p, vector<vector<long long> > &up,
           vector<vector<long long> > &down, vector<long long> &ans,
           vector<vector<long long> > &graph, vector<long long> &ans_d) {
  if (p != -1) {
    for (int i = 1; i < up[p].size(); ++i) {
      up[v][i] += up[p][i - 1];
    }
    up[v][0] += up[p].back();
    up[v][min(1, k - 1)]++;
    ans[v] += ans[p];
    ans[v] += up[p][0];
    ans[v]++;
    for (int i = 1; i < down[p].size(); ++i) {
      up[v][i] += down[p][i - 1];
    }
    up[v][0] += down[p].back();
    if (down[p][min(k - 1, 1)]) {
      up[v][min(1, k - 1)] + down[p][0];
    }
    ans[v] += ans_d[p];
    ans[v] += down[p][0];
  }
  for (int i = 0; i < graph[v].size(); ++i) {
    long long to = graph[v][i];
    if (to == p) {
      continue;
    }
    solve(to, v, up, down, ans, graph, ans_d);
    ans_d[v] += ans_d[to];
    ans_d[v] += down[to][0];
    ans_d[v] += 1;
    for (int i = 1; i < down[to].size(); ++i) {
      down[v][i] += down[to][i - 1];
    }
    down[v][0] += down[to].back();
    down[v][min(k - 1, 1)]++;
  }
}
int main() {
  long long n;
  cin >> n >> k;
  vector<vector<long long> > graph(n);
  long long from, to;
  for (int i = 0; i < n - 1; ++i) {
    cin >> from >> to;
    graph[from - 1].push_back(to - 1);
    graph[to - 1].push_back(from - 1);
  }
  vector<vector<long long> > up(n, vector<long long>(k)),
      down(n, vector<long long>(k));
  vector<long long> ans(n), ans_d(n);
  solve(0, -1, up, down, ans, graph, ans_d);
  long long result = 0;
  for (int i = 0; i < n; ++i) {
    result += ans[i];
  }
  cout << result;
}

```