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.
#include <bits/stdc++.h>
using namespace std;
int k;
class Solution {
  int n;
  long long ans = 0;
  struct DP {
    long long sum = 0;
    int cnt[5] = {0};
    DP() { cnt[0] = 1; }
    int count() const { return accumulate(cnt, cnt + k, 0); }
    DP &operator+=(const DP &oth) {
      sum += oth.sum;
      for (int i = 0; i < k; ++i) cnt[i] += oth.cnt[i];
      return *this;
    }
    void shift() {
      rotate(cnt, cnt + k - 1, cnt + k);
      sum += cnt[0];
    }
    long long operator*(const DP &oth) {
      long long ret = sum * oth.count() + oth.sum * count();
      for (int i = 0; i < k; ++i) {
        for (int j = 0; j < k; ++j) {
          ret += 1ll * (i + j + k - 1) / k * cnt[i] * oth.cnt[j];
        }
      }
      return ret;
    }
  };
  vector<vector<int>> g;
  DP dfs(int u, int p) {
    DP ret;
    for (int v : g[u])
      if (v != p) {
        auto cur = dfs(v, u);
        ans += ret * cur;
        ret += cur;
      }
    ret.shift();
    return ret;
  }

 public:
  void run() {
    cin >> n >> k;
    g.resize(n);
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      --a, --b;
      g[a].push_back(b), g[b].push_back(a);
    }
    auto ass = dfs(0, 0);
    cout << ans << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Solution().run();
}
