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 K;
vector<vector<pair<long long, int>>> res, true_res;
void dfs_1(int v, vector<bool> &used, const vector<vector<int>> &graph) {
  used[v] = true;
  res[v][K - 1].second = 1;
  for (int to : graph[v]) {
    if (used[to]) continue;
    dfs_1(to, used, graph);
    res[v][0].first += res[to][K - 1].first + res[to][K - 1].second;
    res[v][0].second += res[to][K - 1].second;
    for (int i = 1; i != K; ++i) {
      res[v][i].first += res[to][i - 1].first;
      res[v][i].second += res[to][i - 1].second;
    }
  }
}
void dfs_2(int v, int from, vector<bool> &used,
           const vector<vector<int>> &graph) {
  used[v] = true;
  auto father_res = true_res[from];
  father_res[0].first -= res[v][K - 1].first + res[v][K - 1].second;
  father_res[0].second -= res[v][K - 1].second;
  for (int i = 1; i != K; ++i) {
    father_res[i].first -= res[v][i - 1].first;
    father_res[i].second -= res[v][i - 1].second;
  }
  true_res[v] = res[v];
  true_res[v][0].first += father_res[K - 1].first + father_res[K - 1].second;
  true_res[v][0].second += father_res[K - 1].second;
  for (int i = 1; i != K; ++i) {
    true_res[v][i].first += father_res[i - 1].first;
    true_res[v][i].second += father_res[i - 1].second;
  }
  for (int to : graph[v]) {
    if (used[to]) continue;
    dfs_2(to, v, used, graph);
  }
}
int main() {
  int n;
  cin >> n >> K;
  vector<vector<int>> graph(n);
  for (int i = 0; i != n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  vector<bool> used(n, false);
  res.resize(n, vector<pair<long long, int>>(K));
  dfs_1(0, used, graph);
  true_res.resize(n);
  true_res[0] = res[0];
  used.assign(n, false);
  used[0] = true;
  for (int to : graph[0]) {
    dfs_2(to, 0, used, graph);
  }
  long long answer = 0;
  for (int i = 0; i != n; ++i) {
    for (int j = 0; j != K; ++j) {
      answer += true_res[i][j].first;
    }
  }
  cout << answer / 2 << '\n';
  return 0;
}
