#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 7;
int n, k;
long long sum[MAX];
long long cnt[MAX][10];
long long total_cnt[MAX][10];
long long sz[MAX], total[MAX];
vector<int> adj[MAX];
void dfs1(int i, int p = -1) {
  sz[i] = 1;
  cnt[i][0] = 1;
  for (auto j : adj[i])
    if (j != p) {
      dfs1(j, i);
      sz[i] += sz[j];
      sum[i] += sz[j] + sum[j];
      for (int p = 0; p < k; p++) cnt[i][(p + 1) % k] += cnt[j][p];
    }
}
void dfs2(int i, int p = -1) {
  for (auto j : adj[i])
    if (j != p) {
      long long temp = total[i] - sum[j] - sz[j];
      total[j] = sum[j] + temp + (n - sz[j]);
      for (int p = 0; p < k; p++) {
        temp = total_cnt[i][p] - cnt[j][(p - 1 + k) % k];
        total_cnt[j][(p + 1) % k] = temp + cnt[j][(p + 1) % k];
      }
      dfs2(j, i);
    }
}
long long func() {
  dfs1(1);
  total[1] = sum[1];
  for (int i = 0; i < k; i++) total_cnt[1][i] = cnt[1][i];
  dfs2(1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (k == 1) {
      ans += total[i];
      continue;
    }
    long long temp = total[i];
    for (int p = 1; p < k; p++) {
      temp -= (long long)p * total_cnt[i][p];
    }
    assert(temp % k == 0);
    ans += temp / k;
    for (int p = 1; p < k; p++) ans += total_cnt[i][p];
  }
  ans = ans / 2;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cout << func() << 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;
const int MAX = 2e5 + 7;
int n, k;
long long sum[MAX];
long long cnt[MAX][10];
long long total_cnt[MAX][10];
long long sz[MAX], total[MAX];
vector<int> adj[MAX];
void dfs1(int i, int p = -1) {
  sz[i] = 1;
  cnt[i][0] = 1;
  for (auto j : adj[i])
    if (j != p) {
      dfs1(j, i);
      sz[i] += sz[j];
      sum[i] += sz[j] + sum[j];
      for (int p = 0; p < k; p++) cnt[i][(p + 1) % k] += cnt[j][p];
    }
}
void dfs2(int i, int p = -1) {
  for (auto j : adj[i])
    if (j != p) {
      long long temp = total[i] - sum[j] - sz[j];
      total[j] = sum[j] + temp + (n - sz[j]);
      for (int p = 0; p < k; p++) {
        temp = total_cnt[i][p] - cnt[j][(p - 1 + k) % k];
        total_cnt[j][(p + 1) % k] = temp + cnt[j][(p + 1) % k];
      }
      dfs2(j, i);
    }
}
long long func() {
  dfs1(1);
  total[1] = sum[1];
  for (int i = 0; i < k; i++) total_cnt[1][i] = cnt[1][i];
  dfs2(1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (k == 1) {
      ans += total[i];
      continue;
    }
    long long temp = total[i];
    for (int p = 1; p < k; p++) {
      temp -= (long long)p * total_cnt[i][p];
    }
    assert(temp % k == 0);
    ans += temp / k;
    for (int p = 1; p < k; p++) ans += total_cnt[i][p];
  }
  ans = ans / 2;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cout << func() << endl;
  return 0;
}

```