#include <bits/stdc++.h>
using namespace std;
vector<int> a[200005];
long long f[200005][15], f1[200005][15], d[200005], d1[200005];
long long n, k, kq;
void nhap() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  int x, y;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
}
void dfs(int x, int p) {
  ++f[x][0];
  for (int i = 0; i < a[x].size(); ++i)
    if (a[x][i] != p) {
      dfs(a[x][i], x);
      for (int j = 1; j <= k; ++j) f[x][j] += f[a[x][i]][j - 1];
      f[x][1] += f[a[x][i]][k];
      d[x] += d[a[x][i]];
    }
  d[x] += f[x][1];
  kq += d[x];
}
void dfs1(int x, int p) {
  for (int i = 0; i < a[x].size(); ++i)
    if (a[x][i] != p) {
      for (int j = 1; j <= k; ++j)
        if (k > 1)
          f1[a[x][i]][j] = f1[x][j - 1] + f[x][j - 1] - f[a[x][i]][j - 2];
        else
          f1[a[x][i]][j] = f1[x][j - 1] + f[x][j - 1];
      if (k > 1)
        f1[a[x][i]][1] += (f1[x][k] + f[x][k] - f[a[x][i]][k - 1]);
      else
        f1[a[x][i]][1] +=
            f1[x][k] + f[x][k] - f[a[x][i]][k] - f[a[x][i]][k - 1];
      f1[a[x][i]][2] -= f[a[x][i]][k];
      d1[a[x][i]] = d1[x] + d[x] - d[a[x][i]] - f[a[x][i]][k] - f[a[x][i]][0];
      d1[a[x][i]] += f1[a[x][i]][1];
      kq += d1[a[x][i]];
      dfs1(a[x][i], x);
    }
}
int main() {
  nhap();
  dfs(1, 0);
  dfs1(1, 0);
  cout << kq / 2;
  return 0;
}

### Prompt

Please formulate a Cpp solution 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>
using namespace std;
vector<int> a[200005];
long long f[200005][15], f1[200005][15], d[200005], d1[200005];
long long n, k, kq;
void nhap() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  int x, y;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
}
void dfs(int x, int p) {
  ++f[x][0];
  for (int i = 0; i < a[x].size(); ++i)
    if (a[x][i] != p) {
      dfs(a[x][i], x);
      for (int j = 1; j <= k; ++j) f[x][j] += f[a[x][i]][j - 1];
      f[x][1] += f[a[x][i]][k];
      d[x] += d[a[x][i]];
    }
  d[x] += f[x][1];
  kq += d[x];
}
void dfs1(int x, int p) {
  for (int i = 0; i < a[x].size(); ++i)
    if (a[x][i] != p) {
      for (int j = 1; j <= k; ++j)
        if (k > 1)
          f1[a[x][i]][j] = f1[x][j - 1] + f[x][j - 1] - f[a[x][i]][j - 2];
        else
          f1[a[x][i]][j] = f1[x][j - 1] + f[x][j - 1];
      if (k > 1)
        f1[a[x][i]][1] += (f1[x][k] + f[x][k] - f[a[x][i]][k - 1]);
      else
        f1[a[x][i]][1] +=
            f1[x][k] + f[x][k] - f[a[x][i]][k] - f[a[x][i]][k - 1];
      f1[a[x][i]][2] -= f[a[x][i]][k];
      d1[a[x][i]] = d1[x] + d[x] - d[a[x][i]] - f[a[x][i]][k] - f[a[x][i]][0];
      d1[a[x][i]] += f1[a[x][i]][1];
      kq += d1[a[x][i]];
      dfs1(a[x][i], x);
    }
}
int main() {
  nhap();
  dfs(1, 0);
  dfs1(1, 0);
  cout << kq / 2;
  return 0;
}

```