#include <bits/stdc++.h>
using namespace std;
int n, k;
struct subtree {
  unsigned long long waystoroot[5], sumtoroot[5];
  unsigned long long inside;
  subtree() {
    inside = 0;
    for (int i = 0; i < k; i++) {
      waystoroot[i] = sumtoroot[i] = 0;
    }
  }
};
vector<int> tree[200001];
void merge(subtree &first, subtree &second) {
  first.inside += second.inside;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      first.inside += first.waystoroot[i] * second.sumtoroot[j] +
                      first.sumtoroot[i] * second.waystoroot[j];
      if (i + j >= k)
        first.inside -= (first.waystoroot[i] * second.waystoroot[j]);
    }
    first.sumtoroot[i] += second.sumtoroot[i];
    first.waystoroot[i] += second.waystoroot[i];
  }
}
subtree dfs(int cur, int par) {
  vector<subtree> subvc;
  for (auto &x : tree[cur]) {
    if (x == par) continue;
    subvc.push_back(dfs(x, cur));
  }
  if (subvc.empty()) return subtree();
  for (auto &x : subvc) {
    long long tmpways[5], tmpsum[5];
    for (int i = 1; i < k; i++) {
      x.inside += x.sumtoroot[i];
      tmpways[i - 1] = x.waystoroot[i];
      tmpsum[i - 1] = x.sumtoroot[i];
    }
    x.inside += x.sumtoroot[0];
    tmpways[k - 1] = x.waystoroot[0];
    tmpsum[k - 1] = x.sumtoroot[0] + x.waystoroot[0];
    tmpsum[k - 1]++, tmpways[k - 1]++;
    for (int i = 0; i < k; i++)
      x.sumtoroot[i] = tmpsum[i], x.waystoroot[i] = tmpways[i];
  }
  for (int i = 1; i < subvc.size(); i++) merge(subvc[0], subvc[i]);
  return subvc[0];
}
int main() {
  int a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  subtree subrez = dfs(1, 1);
  unsigned long long rez = subrez.inside;
  for (int i = 0; i < k; i++) rez += subrez.sumtoroot[i];
  cout << rez << endl;
  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>
using namespace std;
int n, k;
struct subtree {
  unsigned long long waystoroot[5], sumtoroot[5];
  unsigned long long inside;
  subtree() {
    inside = 0;
    for (int i = 0; i < k; i++) {
      waystoroot[i] = sumtoroot[i] = 0;
    }
  }
};
vector<int> tree[200001];
void merge(subtree &first, subtree &second) {
  first.inside += second.inside;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      first.inside += first.waystoroot[i] * second.sumtoroot[j] +
                      first.sumtoroot[i] * second.waystoroot[j];
      if (i + j >= k)
        first.inside -= (first.waystoroot[i] * second.waystoroot[j]);
    }
    first.sumtoroot[i] += second.sumtoroot[i];
    first.waystoroot[i] += second.waystoroot[i];
  }
}
subtree dfs(int cur, int par) {
  vector<subtree> subvc;
  for (auto &x : tree[cur]) {
    if (x == par) continue;
    subvc.push_back(dfs(x, cur));
  }
  if (subvc.empty()) return subtree();
  for (auto &x : subvc) {
    long long tmpways[5], tmpsum[5];
    for (int i = 1; i < k; i++) {
      x.inside += x.sumtoroot[i];
      tmpways[i - 1] = x.waystoroot[i];
      tmpsum[i - 1] = x.sumtoroot[i];
    }
    x.inside += x.sumtoroot[0];
    tmpways[k - 1] = x.waystoroot[0];
    tmpsum[k - 1] = x.sumtoroot[0] + x.waystoroot[0];
    tmpsum[k - 1]++, tmpways[k - 1]++;
    for (int i = 0; i < k; i++)
      x.sumtoroot[i] = tmpsum[i], x.waystoroot[i] = tmpways[i];
  }
  for (int i = 1; i < subvc.size(); i++) merge(subvc[0], subvc[i]);
  return subvc[0];
}
int main() {
  int a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  subtree subrez = dfs(1, 1);
  unsigned long long rez = subrez.inside;
  for (int i = 0; i < k; i++) rez += subrez.sumtoroot[i];
  cout << rez << endl;
  return 0;
}

```