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 n, k;
vector<int> v[200010];
long long cnt = 0, ch[200010][10], sm[200010][10], tmp[10];
void dfs(int x, int p) {
  for (auto e : v[x]) {
    if (e != p) {
      dfs(e, x);
      for (int i = 0; i < k; ++i) {
        ch[x][(i + 1) % k] += ch[e][i];
        sm[x][(i + 1) % k] += sm[e][i];
      }
      ++ch[x][1 % k];
      sm[x][1 % k] += ch[e][0] + 1;
    }
  }
  for (auto e : v[x]) {
    if (e != p) {
      for (int i = 0; i < k; ++i) {
        for (int j = 0; j < k; ++j) {
          long long tmp1 = sm[e][(i + k - 1) % k],
                    tmp2 = sm[x][j] - sm[e][(j + k - 1) % k];
          long long ch1 = ch[e][(i + k - 1) % k],
                    ch2 = ch[x][j] - ch[e][(j + k - 1) % k];
          if (i == 1 % k) {
            tmp1 += ch[e][0] + 1;
            ++ch1;
          }
          if (j == 1 % k) {
            tmp2 -= ch[e][0] + 1;
            --ch2;
          }
          cnt += tmp1 * ch2 + tmp2 * ch1;
          if (i && j && i + j <= k) {
            cnt -= ch1 * ch2;
          }
        }
      }
    }
  }
}
int main() {
  cin >> n >> k;
  memset(ch, 0, sizeof ch);
  memset(sm, 0, sizeof sm);
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  cnt >>= 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) {
      cnt += sm[i][j];
    }
  }
  cout << cnt << endl;
  return 0;
}
