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;
const int MXN = 200100;
vector<int> ed[MXN];
int was[MXN];
int k;
long long value(const vector<int>& dcnt) { return dcnt[0]; }
vector<int> calc_new(vector<int> a) {
  rotate(a.begin(), a.begin() + a.size() - 1, a.end());
  return a;
}
vector<int> add_vec(const vector<int>& a, const vector<int>& b) {
  int n = a.size();
  vector<int> res(n);
  for (int i = (0); i < (n); i++) {
    res[i] = a[i] + b[i];
  }
  return res;
}
long long res = 0;
pair<vector<int>, long long> dfs(int v, int p, long long sum,
                                 vector<int> dcnt) {
  if (k > 1) {
    dcnt[1]++;
  } else {
    dcnt[0]++;
  }
  sum++;
  res += sum;
  int sz = ed[v].size();
  vector<int> diff(k);
  long long val_diff = 0;
  for (int i = (0); i < (sz); i++) {
    int to = ed[v][i];
    if (to == p) {
      continue;
    }
    pair<vector<int>, long long> dd =
        dfs(to, v, val_diff + sum + value(dcnt) + value(diff),
            add_vec(calc_new(dcnt), calc_new(diff)));
    val_diff += value(dd.first);
    val_diff += dd.second;
    dd.first = calc_new(dd.first);
    diff = add_vec(diff, dd.first);
  }
  diff[0]++;
  return {diff, val_diff};
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    ed[a].push_back(b);
    ed[b].push_back(a);
  }
  vector<int> cnt(k);
  if (k > 1) {
    cnt[1] = -1;
  } else {
    cnt[0] = -1;
  }
  dfs(0, -1, -1, cnt);
  cout << res << endl;
}
