#include <bits/stdc++.h>
using namespace std;
vector<int> v[200010];
int subtree[200010][10], tree[200010][10], p[10], D[200010], odl[200010], n, k,
    a, b;
bool bylem[200010];
long long sumka, suma, add, x;
void DFSdp(int w) {
  bylem[w] = true;
  sumka += odl[w];
  subtree[w][0] = D[w] = 1;
  for (int i = 0; i < v[w].size(); i++) {
    if (!bylem[v[w][i]]) {
      odl[v[w][i]] = odl[w] + 1;
      DFSdp(v[w][i]);
      D[w] += D[v[w][i]];
      for (int j = 0; j < k; j++) {
        subtree[w][j] += subtree[v[w][i]][(j + k - 1) % k];
      }
    }
  }
}
void DFSsum(int w, long long sum) {
  suma += sum;
  for (int i = 1; i < k; i++) {
    add += tree[w][i] * (k - i);
  }
  bylem[w] = true;
  for (int i = 0; i < v[w].size(); i++) {
    if (!bylem[v[w][i]]) {
      for (int j = 0; j < k; j++) {
        p[j] = tree[w][j] - subtree[v[w][i]][(j + k - 1) % k];
      }
      for (int j = 0; j < k; j++) {
        tree[v[w][i]][(j + 1) % k] = p[j] + subtree[v[w][i]][(j + 1) % k];
      }
      DFSsum(v[w][i], sum - D[v[w][i]] + n - D[v[w][i]]);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b), v[b].push_back(a);
  }
  DFSdp(1);
  for (int i = 1; i <= n; i++) {
    bylem[i] = false;
  }
  for (int i = 0; i < k; i++) {
    tree[1][i] = subtree[1][i];
  }
  DFSsum(1, sumka);
  printf("%lld", ((suma >> 1) + (add >> 1)) / k);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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> v[200010];
int subtree[200010][10], tree[200010][10], p[10], D[200010], odl[200010], n, k,
    a, b;
bool bylem[200010];
long long sumka, suma, add, x;
void DFSdp(int w) {
  bylem[w] = true;
  sumka += odl[w];
  subtree[w][0] = D[w] = 1;
  for (int i = 0; i < v[w].size(); i++) {
    if (!bylem[v[w][i]]) {
      odl[v[w][i]] = odl[w] + 1;
      DFSdp(v[w][i]);
      D[w] += D[v[w][i]];
      for (int j = 0; j < k; j++) {
        subtree[w][j] += subtree[v[w][i]][(j + k - 1) % k];
      }
    }
  }
}
void DFSsum(int w, long long sum) {
  suma += sum;
  for (int i = 1; i < k; i++) {
    add += tree[w][i] * (k - i);
  }
  bylem[w] = true;
  for (int i = 0; i < v[w].size(); i++) {
    if (!bylem[v[w][i]]) {
      for (int j = 0; j < k; j++) {
        p[j] = tree[w][j] - subtree[v[w][i]][(j + k - 1) % k];
      }
      for (int j = 0; j < k; j++) {
        tree[v[w][i]][(j + 1) % k] = p[j] + subtree[v[w][i]][(j + 1) % k];
      }
      DFSsum(v[w][i], sum - D[v[w][i]] + n - D[v[w][i]]);
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b), v[b].push_back(a);
  }
  DFSdp(1);
  for (int i = 1; i <= n; i++) {
    bylem[i] = false;
  }
  for (int i = 0; i < k; i++) {
    tree[1][i] = subtree[1][i];
  }
  DFSsum(1, sumka);
  printf("%lld", ((suma >> 1) + (add >> 1)) / k);
  return 0;
}

```