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<vector<long long> > graph;
vector<bool> visited;
long long ans;
long long *getArray(long long size) {
  long long *arr = new long long[size];
  for (long long i = 0; i < size; i++) arr[i] = 0;
  return arr;
}
void merge(long long *childarr, long long *parentarr, long long lcadepth,
           long long k) {
  for (long long i = 0; i < k; i++) {
    for (long long j = 0; j < k; j++) {
      long long a = i + j;
      long long b = 2 * lcadepth;
      long long rem = ((a - b) % k + k) % k;
      ans += (((k - rem) % k + k) % k) * childarr[i] * parentarr[j];
    }
  }
}
void updateArrSum(long long *arr, long long *childArr, long long k) {
  for (long long i = 0; i < k; i++) {
    arr[i] += childArr[i];
  }
}
long long *dfs(long long current_node, long long &subtreesize, long long k,
               long long n, long long depth) {
  long long *arr = getArray(k);
  long long subtreesize_local = 1;
  arr[depth % k] = 1;
  for (long long i = 0; i < graph[current_node].size(); i++) {
    long long child = graph[current_node][i];
    if (!visited[child]) {
      visited[child] = true;
      long long *childarr = dfs(child, subtreesize_local, k, n, depth + 1);
      merge(childarr, arr, depth, k);
      updateArrSum(arr, childarr, k);
    }
  }
  subtreesize += subtreesize_local;
  ans += subtreesize_local * (n - subtreesize_local);
  return arr;
}
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n + 1; i++) {
    graph.push_back(vector<long long>());
    visited.push_back(false);
  }
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  ans = 0;
  visited[1] = true;
  long long subtreesize = 0;
  dfs(1, subtreesize, k, n, 0);
  cout << ans / k;
  return 0;
}
