#include <bits/stdc++.h>
using namespace std;
typedef struct Node {
  long long ans, step;
  long long rmn[5];
  int cnt;
  vector<int> edge;
} Node;
int n, k;
Node node[200001];
bool visited[200001];
void dfs(int v) {
  int i, j, l, u, t;
  visited[v] = true;
  node[v].ans = node[v].step = 0;
  node[v].cnt = 0;
  node[v].rmn[0] = node[v].rmn[1] = node[v].rmn[2] = node[v].rmn[3] =
      node[v].rmn[4] = 0;
  for (i = 0; i < node[v].edge.size(); i++) {
    u = node[v].edge[i];
    if (visited[u]) continue;
    dfs(u);
    node[v].ans += node[u].cnt * node[v].step + node[v].cnt * node[u].step;
    node[v].ans += node[u].step;
    for (j = 1; j < k; j++)
      for (l = 1; l < k; l++) {
        if (j + l <= k) node[v].ans -= node[u].rmn[j] * node[v].rmn[l];
      }
    for (j = 0; j < k; j++) node[v].rmn[j] += node[u].rmn[j];
    node[v].step += node[u].step;
    node[v].cnt += node[u].cnt;
    node[v].ans += node[u].ans;
  }
  node[v].cnt++;
  node[v].rmn[0]++;
  node[v].step += node[v].rmn[0];
  node[v].ans += node[v].rmn[0];
  t = node[v].rmn[k - 1];
  for (i = k - 1; i >= 1; i--) node[v].rmn[i] = node[v].rmn[i - 1];
  node[v].rmn[0] = t;
}
int main() {
  int i, u, v;
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    cin >> u >> v;
    node[u].edge.push_back(v);
    node[v].edge.push_back(u);
  }
  dfs(1);
  cout << node[1].ans - node[1].step << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
typedef struct Node {
  long long ans, step;
  long long rmn[5];
  int cnt;
  vector<int> edge;
} Node;
int n, k;
Node node[200001];
bool visited[200001];
void dfs(int v) {
  int i, j, l, u, t;
  visited[v] = true;
  node[v].ans = node[v].step = 0;
  node[v].cnt = 0;
  node[v].rmn[0] = node[v].rmn[1] = node[v].rmn[2] = node[v].rmn[3] =
      node[v].rmn[4] = 0;
  for (i = 0; i < node[v].edge.size(); i++) {
    u = node[v].edge[i];
    if (visited[u]) continue;
    dfs(u);
    node[v].ans += node[u].cnt * node[v].step + node[v].cnt * node[u].step;
    node[v].ans += node[u].step;
    for (j = 1; j < k; j++)
      for (l = 1; l < k; l++) {
        if (j + l <= k) node[v].ans -= node[u].rmn[j] * node[v].rmn[l];
      }
    for (j = 0; j < k; j++) node[v].rmn[j] += node[u].rmn[j];
    node[v].step += node[u].step;
    node[v].cnt += node[u].cnt;
    node[v].ans += node[u].ans;
  }
  node[v].cnt++;
  node[v].rmn[0]++;
  node[v].step += node[v].rmn[0];
  node[v].ans += node[v].rmn[0];
  t = node[v].rmn[k - 1];
  for (i = k - 1; i >= 1; i--) node[v].rmn[i] = node[v].rmn[i - 1];
  node[v].rmn[0] = t;
}
int main() {
  int i, u, v;
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    cin >> u >> v;
    node[u].edge.push_back(v);
    node[v].edge.push_back(u);
  }
  dfs(1);
  cout << node[1].ans - node[1].step << endl;
  return 0;
}

```