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 test;
long long n, k;
long long a, b;
vector<long long> g[200007];
long long dp[200007][10];
long long sub[200007], lel[200007];
long long res;
void dfs(long long v, long long p) {
  sub[v] = 1;
  dp[v][0] = 1;
  for (auto it : g[v]) {
    if (it != p) {
      dfs(it, v);
      res += sub[it] * lel[v];
      res += sub[v] * lel[it];
      sub[v] += sub[it];
      lel[v] += lel[it];
      lel[v] += dp[it][k - 1];
      for (long long i = 0; i < (k); ++i) {
        for (long long j = 0; j < (k); ++j) {
          if (i + j + 1 > k) {
            res += dp[it][i] * dp[v][j];
          }
          res += dp[it][i] * dp[v][j];
        }
      }
      for (long long i = 0; i < (k); ++i) {
        dp[v][(i + 1) % k] += dp[it][i];
      }
    }
  }
  return;
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i <= (n - 1); ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, 1);
  cout << res;
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
