#include <bits/stdc++.h>
using namespace std;
struct rast {
  long long int r[6], k;
  rast() {
    k = 0;
    for (int i = 1; i < 6; i++) {
      r[i] = 0;
    }
    r[0] = 1;
  }
};
int n, k;
vector<vector<int> > a;
long long int ans;
rast r[202345];
void Add(int v, int to) {
  for (int i = 0; i <= k; i++) {
    ans += (r[v].k * r[to].r[i]) + (r[to].k * r[v].r[i]);
    for (int u = 0; u <= k; u++) {
      if (i + u < k)
        ans += r[v].r[i] * r[to].r[u];
      else if (i + u < k * 2)
        ans += 2 * r[v].r[i] * r[to].r[u];
      else
        ans += 3 * r[v].r[i] * r[to].r[u];
    }
  }
  for (int i = 0; i < k; i++) {
    r[v].r[i + 1] += r[to].r[i];
  }
  r[v].r[1] += r[to].r[k];
  r[v].k += r[to].k + r[to].r[k];
}
void dfs(int v, int from) {
  for (int i = 0; i < a[v].size(); i++) {
    int to = a[v][i];
    if (to == from) continue;
    dfs(to, v);
    Add(v, to);
  }
}
int main() {
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(0, -1);
  cout << ans << "\n";
  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;
struct rast {
  long long int r[6], k;
  rast() {
    k = 0;
    for (int i = 1; i < 6; i++) {
      r[i] = 0;
    }
    r[0] = 1;
  }
};
int n, k;
vector<vector<int> > a;
long long int ans;
rast r[202345];
void Add(int v, int to) {
  for (int i = 0; i <= k; i++) {
    ans += (r[v].k * r[to].r[i]) + (r[to].k * r[v].r[i]);
    for (int u = 0; u <= k; u++) {
      if (i + u < k)
        ans += r[v].r[i] * r[to].r[u];
      else if (i + u < k * 2)
        ans += 2 * r[v].r[i] * r[to].r[u];
      else
        ans += 3 * r[v].r[i] * r[to].r[u];
    }
  }
  for (int i = 0; i < k; i++) {
    r[v].r[i + 1] += r[to].r[i];
  }
  r[v].r[1] += r[to].r[k];
  r[v].k += r[to].k + r[to].r[k];
}
void dfs(int v, int from) {
  for (int i = 0; i < a[v].size(); i++) {
    int to = a[v][i];
    if (to == from) continue;
    dfs(to, v);
    Add(v, to);
  }
}
int main() {
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(0, -1);
  cout << ans << "\n";
  return 0;
}

```