#include <bits/stdc++.h>
const int mod = 1e9 + 7;
std::vector<int> e[200001];
long long q[200001][6], n, k;
long long f[200001][6];
long long res;
bool vis[200001];
void DFS(int s) {
  vis[s] = true;
  q[s][0] = 1;
  int u;
  for (int i = 0; i < e[s].size(); ++i) {
    if (vis[e[s][i]]) continue;
    DFS(e[s][i]);
    u = e[s][i];
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        res += q[s][i] * f[u][j] + f[s][i] * q[u][j] +
               q[s][i] * q[u][j] * ((i + j + 1) / k + ((i + j + 1) % k > 0));
      }
    }
    for (int i = 0; i < k; ++i) {
      q[s][i] += q[u][(k + i - 1) % k];
      f[s][i] += f[u][(k + i - 1) % k] + q[u][(k + i - 1) % k] * (!i);
    }
  }
}
int main() {
  std::cin >> n >> k;
  int l, r;
  for (int i = 1; i < n; ++i) {
    std::cin >> l >> r;
    e[l].push_back(r);
    e[r].push_back(l);
  }
  DFS(1);
  std::cout << res;
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int mod = 1e9 + 7;
std::vector<int> e[200001];
long long q[200001][6], n, k;
long long f[200001][6];
long long res;
bool vis[200001];
void DFS(int s) {
  vis[s] = true;
  q[s][0] = 1;
  int u;
  for (int i = 0; i < e[s].size(); ++i) {
    if (vis[e[s][i]]) continue;
    DFS(e[s][i]);
    u = e[s][i];
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        res += q[s][i] * f[u][j] + f[s][i] * q[u][j] +
               q[s][i] * q[u][j] * ((i + j + 1) / k + ((i + j + 1) % k > 0));
      }
    }
    for (int i = 0; i < k; ++i) {
      q[s][i] += q[u][(k + i - 1) % k];
      f[s][i] += f[u][(k + i - 1) % k] + q[u][(k + i - 1) % k] * (!i);
    }
  }
}
int main() {
  std::cin >> n >> k;
  int l, r;
  for (int i = 1; i < n; ++i) {
    std::cin >> l >> r;
    e[l].push_back(r);
    e[r].push_back(l);
  }
  DFS(1);
  std::cout << res;
  return 0;
}

```