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;
long long ans = 0;
vector<int> v[200005];
long long total[200005], cnt[200005][5], k;
void calc_one(int s, int pr, int c) {
  cnt[s][0] = 1;
  for (auto g : v[s]) {
    if (g != pr) {
      calc_one(g, s, c + 1);
      total[s] += total[g];
      total[s] += cnt[g][k - 1];
      for (int i = 1; i < k; ++i) {
        cnt[s][i] += cnt[g][i - 1];
      }
      cnt[s][0] += cnt[g][k - 1];
    }
  }
}
void dfs(int s, int pr, long long ptotal, vector<long long> pcnt) {
  ans = ans + total[s] + ptotal;
  for (int i = 1; i < k; ++i) {
    ans = ans + cnt[s][i] + pcnt[i];
  }
  for (auto g : v[s]) {
    if (g != pr) {
      vector<long long> tcnt(k);
      long long ttotal = 0;
      ttotal = ptotal + total[s] - total[g];
      ttotal -= cnt[g][k - 1];
      for (int i = 1; i < k; ++i) {
        tcnt[i] = pcnt[i] + cnt[s][i] - cnt[g][i - 1];
      }
      tcnt[0] = pcnt[0] + cnt[s][0] - cnt[g][k - 1];
      ttotal += tcnt[k - 1];
      long long tmp = tcnt[k - 1];
      for (int i = k - 1; i >= 1; --i) {
        tcnt[i] = tcnt[i - 1];
      }
      tcnt[0] = tmp;
      dfs(g, s, ttotal, tcnt);
    }
  }
}
int main() {
  int n, x, y;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  memset(cnt, 0, sizeof(cnt));
  calc_one(1, 0, 0);
  vector<long long> pcnt(k);
  dfs(1, 0, 0, pcnt);
  cout << ans / 2;
}
