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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int N = 2e5 + 7;
long long i, j, n, m, k, used[N], d[N][6], sum[N][6], ans = 0, dd[6], ss[6], kk;
vector<vector<long long> > v(N);
void dfs(int x) {
  used[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    long long to = v[x][i];
    if (used[to] == 1) continue;
    dfs(to);
    for (int j = 0; j <= m; j++) ss[j] = sum[x][j], dd[j] = d[x][j];
    for (int j = 0; j <= m; j++) {
      for (int g = 0; g <= m; g++) {
        k = (j + g + 1) / m + (((j + g + 1) % m) != 0);
        kk = (j != 0) + (g != 0);
        ans += d[to][g] * ss[j] + dd[j] * (sum[to][g] + (k - kk) * d[to][g]);
      }
      ans += sum[to][j] + d[to][j] * (j / m);
      if (j == m) {
        sum[x][(j + 2) % (m + 1)] += sum[to][j] + d[to][j] * (j / m);
        d[x][(j + 2) % (m + 1)] += d[to][j];
      } else {
        sum[x][(j + 1) % (m + 1)] += sum[to][j] + d[to][j] * (j / m);
        d[x][(j + 1) % (m + 1)] += d[to][j];
      }
    }
    sum[x][1]++;
    ans++;
  }
  d[x][0]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0);
  cout << ans << endl;
  return 0;
}
