#include <bits/stdc++.h>
const long long mod = 1000000007L;
using namespace std;
int n, k;
std::vector<int> vec[200010];
long long dp1[10][200010], dp2[10][200010];
long long sub_ans[200010], ans[200010];
void dfs1(int u, int p) {
  dp1[0][u] = 1;
  for (int i : vec[u]) {
    if (i != p) {
      dfs1(i, u);
      for (int j = 0; j < k; j++) {
        dp1[j][u] += dp1[(j - 1 + k) % k][i];
      }
      sub_ans[u] += sub_ans[i] + dp1[0][i];
    }
  }
}
void dfs2(int u, int p) {
  if (p != 0) {
    for (int i = 0; i < k; i++) {
      dp2[i][u] = dp2[(i - 1 + k) % k][p] - dp1[(i - 2 + k) % k][u] + dp1[i][u];
    }
    ans[u] = ans[p] - dp1[0][u] + dp2[0][p] - dp1[k - 1][u];
  }
  for (int i : vec[u]) {
    if (i != p) dfs2(i, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  dfs1(1, 0);
  for (int i = 0; i < k; i++) {
    dp2[i][1] = dp1[i][1];
  }
  ans[1] = sub_ans[1];
  dfs2(1, 0);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += ans[i];
  }
  res = res / 2;
  cout << res << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long mod = 1000000007L;
using namespace std;
int n, k;
std::vector<int> vec[200010];
long long dp1[10][200010], dp2[10][200010];
long long sub_ans[200010], ans[200010];
void dfs1(int u, int p) {
  dp1[0][u] = 1;
  for (int i : vec[u]) {
    if (i != p) {
      dfs1(i, u);
      for (int j = 0; j < k; j++) {
        dp1[j][u] += dp1[(j - 1 + k) % k][i];
      }
      sub_ans[u] += sub_ans[i] + dp1[0][i];
    }
  }
}
void dfs2(int u, int p) {
  if (p != 0) {
    for (int i = 0; i < k; i++) {
      dp2[i][u] = dp2[(i - 1 + k) % k][p] - dp1[(i - 2 + k) % k][u] + dp1[i][u];
    }
    ans[u] = ans[p] - dp1[0][u] + dp2[0][p] - dp1[k - 1][u];
  }
  for (int i : vec[u]) {
    if (i != p) dfs2(i, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  dfs1(1, 0);
  for (int i = 0; i < k; i++) {
    dp2[i][1] = dp1[i][1];
  }
  ans[1] = sub_ans[1];
  dfs2(1, 0);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += ans[i];
  }
  res = res / 2;
  cout << res << endl;
  return 0;
}

```