#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> G[200000 + 5];
long long solve;
long long sub_tree[200000 + 5];
long long sub_tree_mod[200000 + 5][6];
void dfs(int node, int parent, int depth) {
  sub_tree[node] = 1;
  sub_tree_mod[node][depth % k] = 1;
  for (int l = 0; l < G[node].size(); l++) {
    int ady = G[node][l];
    if (ady == parent) continue;
    dfs(ady, node, depth + 1);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        int dist = (i + j + k - (2 * depth) % k) % k;
        int miss = k - dist;
        if (miss == k) miss = 0;
        solve += (long long)miss * sub_tree_mod[node][i] * sub_tree_mod[ady][j];
      }
    }
    for (int i = 0; i < k; i++) {
      sub_tree_mod[node][i] += sub_tree_mod[ady][i];
    }
    sub_tree[node] += sub_tree[ady];
  }
  solve += (long long)sub_tree[node] * (n - sub_tree[node]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0, 0);
  cout << solve / k << '\n';
}

### 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;
int n, k;
vector<int> G[200000 + 5];
long long solve;
long long sub_tree[200000 + 5];
long long sub_tree_mod[200000 + 5][6];
void dfs(int node, int parent, int depth) {
  sub_tree[node] = 1;
  sub_tree_mod[node][depth % k] = 1;
  for (int l = 0; l < G[node].size(); l++) {
    int ady = G[node][l];
    if (ady == parent) continue;
    dfs(ady, node, depth + 1);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        int dist = (i + j + k - (2 * depth) % k) % k;
        int miss = k - dist;
        if (miss == k) miss = 0;
        solve += (long long)miss * sub_tree_mod[node][i] * sub_tree_mod[ady][j];
      }
    }
    for (int i = 0; i < k; i++) {
      sub_tree_mod[node][i] += sub_tree_mod[ady][i];
    }
    sub_tree[node] += sub_tree[ady];
  }
  solve += (long long)sub_tree[node] * (n - sub_tree[node]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0, 0);
  cout << solve / k << '\n';
}

```