#include <bits/stdc++.h>
using namespace std;
struct cww {
  cww() {
    if (1) {
      ios::sync_with_stdio(false);
      cin.tie(0);
    }
  }
} star;
template <typename T>
inline void chmin(T &l, T r) {
  l = min(l, r);
}
template <typename T>
inline void chmax(T &l, T r) {
  l = max(l, r);
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &it : v) is >> it;
  return is;
}
int n, k;
vector<vector<int>> g;
long long res = 0;
tuple<long long, long long, vector<long long>> dfs(int v, int p) {
  long long path_sum = 1;
  long long weight_sum = 0;
  vector<long long> dp(k, 0), nxt(k, 0);
  dp[0] = 1;
  for (auto u : g[v])
    if (u != p) {
      long long sp, sw;
      vector<long long> sd;
      tie(sp, sw, sd) = dfs(u, v);
      for (int i = (0); i < (k); i++)
        for (int j = (0); j < (k); j++) {
          res += ((i + j + k - 1) / k) * dp[i] * sd[j];
        }
      res += weight_sum * sp + sw * path_sum;
      for (int i = (0); i < (k); i++) dp[i] += sd[i];
      weight_sum += sw;
      path_sum += sp;
    }
  weight_sum += dp[k - 1];
  for (int i = (0); i < (k); i++) nxt[(i + 1) % k] = dp[i];
  return tuple<long long, long long, vector<long long>>(path_sum, weight_sum,
                                                        nxt);
}
int main() {
  cin >> n >> k;
  g.resize(n);
  for (int i = (1); i < (n); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  cout << res << 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;
struct cww {
  cww() {
    if (1) {
      ios::sync_with_stdio(false);
      cin.tie(0);
    }
  }
} star;
template <typename T>
inline void chmin(T &l, T r) {
  l = min(l, r);
}
template <typename T>
inline void chmax(T &l, T r) {
  l = max(l, r);
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &it : v) is >> it;
  return is;
}
int n, k;
vector<vector<int>> g;
long long res = 0;
tuple<long long, long long, vector<long long>> dfs(int v, int p) {
  long long path_sum = 1;
  long long weight_sum = 0;
  vector<long long> dp(k, 0), nxt(k, 0);
  dp[0] = 1;
  for (auto u : g[v])
    if (u != p) {
      long long sp, sw;
      vector<long long> sd;
      tie(sp, sw, sd) = dfs(u, v);
      for (int i = (0); i < (k); i++)
        for (int j = (0); j < (k); j++) {
          res += ((i + j + k - 1) / k) * dp[i] * sd[j];
        }
      res += weight_sum * sp + sw * path_sum;
      for (int i = (0); i < (k); i++) dp[i] += sd[i];
      weight_sum += sw;
      path_sum += sp;
    }
  weight_sum += dp[k - 1];
  for (int i = (0); i < (k); i++) nxt[(i + 1) % k] = dp[i];
  return tuple<long long, long long, vector<long long>>(path_sum, weight_sum,
                                                        nxt);
}
int main() {
  cin >> n >> k;
  g.resize(n);
  for (int i = (1); i < (n); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  cout << res << endl;
  return 0;
}

```