#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
const int M = 1e9 + 7;
const double eps = 1e-6;
const double PI(acos(-1.0));
int n, k;
long long tot = 0;
long long f[N][5];
vector<int> g[N];
int sz[N];
void dfs(int u, int p = 0, int dep = 0) {
  f[u][dep % k]++;
  sz[u] = 1;
  for (auto v : g[u])
    if (v - p) {
      dfs(v, u, dep + 1);
      for (int a = 0; a < k; a++)
        for (int b = 0; b < k; b++) {
          int dis = (a + b - 2 * dep + k) % k;
          dis = (k - dis + k) % k;
          tot += dis * f[u][a] * f[v][b];
        }
      sz[u] += sz[v];
      for (int j = 0; j < (k); j++) f[u][j] += f[v][j];
    }
  tot += 1LL * sz[u] * (n - sz[u]);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  cout << tot / k << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 1e6 + 1;
const int M = 1e9 + 7;
const double eps = 1e-6;
const double PI(acos(-1.0));
int n, k;
long long tot = 0;
long long f[N][5];
vector<int> g[N];
int sz[N];
void dfs(int u, int p = 0, int dep = 0) {
  f[u][dep % k]++;
  sz[u] = 1;
  for (auto v : g[u])
    if (v - p) {
      dfs(v, u, dep + 1);
      for (int a = 0; a < k; a++)
        for (int b = 0; b < k; b++) {
          int dis = (a + b - 2 * dep + k) % k;
          dis = (k - dis + k) % k;
          tot += dis * f[u][a] * f[v][b];
        }
      sz[u] += sz[v];
      for (int j = 0; j < (k); j++) f[u][j] += f[v][j];
    }
  tot += 1LL * sz[u] * (n - sz[u]);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  cout << tot / k << endl;
}

```