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;
vector<int> adj[300000];
long long ans;
long long n;
long long k;
long long res[300000][6][2];
void dfs(int v, int p) {
  for (int u : adj[v]) {
    if (u != p) {
      dfs(u, v);
    }
  }
  for (int i = 0; i < k; ++i) {
    for (int u : adj[v]) {
      if (u != p) {
        res[v][i][0] += res[u][(i - 1 + k) % k][0];
        res[v][i][1] += res[u][(i - 1 + k) % k][1] + res[u][(i - 1 + k) % k][0];
      }
    }
  }
  res[v][0][0] += 1;
  for (long long i = 0; i < k; ++i) {
    ans += ((k - i) % k * res[v][i][0] + res[v][i][1]) / k;
  }
  vector<long long> acc0(k, 0);
  vector<long long> acc1(k, 0);
  for (int u : adj[v]) {
    if (u != p) {
      for (long long i = 0; i < k; ++i) {
        for (long long j = 0; j < k; ++j) {
          long long n0 = res[u][i][0] * acc0[j];
          long long rem = (i + j + 2) % k;
          long long n1 =
              2 * n0 + res[u][i][1] * acc0[j] + res[u][i][0] * acc1[j];
          ans += ((k - rem) % k * n0 + n1) / k;
        }
      }
      for (int i = 0; i < k; ++i) {
        acc0[i] += res[u][i][0];
        acc1[i] += res[u][i][1];
      }
    }
  }
}
int main() {
  ios ::sync_with_stdio(0);
  while (cin >> n >> k) {
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      a--;
      b--;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    ans = 0;
    memset(res, 0, sizeof(res));
    dfs(0, -1);
    cout << ans << "\n";
    for (int i = 0; i < n; ++i) {
      adj[i].clear();
    }
  }
  return 0;
}
