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;
using ll = long long;
ll dp_sum[5][200001];
ll dp_no[5][200001];
bool vis[200001];
vector<int> adj[200001];
int k;
ll ans = 0LL;
ll ans2 = 0LL;
void dfs(int v, int par) {
  vis[v] = 1;
  for (auto u : adj[v]) {
    if (u != par) {
      dfs(u, v);
      for (int i = 1; i <= k; i++) {
        dp_sum[i % k][v] += dp_no[i - 1][u] + dp_sum[i - 1][u];
        dp_no[i % k][v] += dp_no[i - 1][u];
      }
    }
  }
  for (auto u : adj[v]) {
    if (u == par) continue;
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        ll no_paths = dp_no[(i + k - 1) % k][u] *
                      (dp_no[j][v] - dp_no[(j + k - 1) % k][u]);
        ans += (dp_sum[(i + k - 1) % k][u] + dp_no[(i + k - 1) % k][u]) *
               (dp_no[j][v] - dp_no[(j + k - 1) % k][u]);
        ans += (dp_sum[j][v] - dp_sum[(j + k - 1) % k][u] -
                dp_no[(j + k - 1) % k][u]) *
               dp_no[(i + k - 1) % k][u];
        ans += ((k + k - i - j) % k) * no_paths;
      }
    }
  }
  for (int i = 0; i < k; i++) {
    ans2 += dp_sum[i][v] + ((k - i) % k) * dp_no[i][v];
  }
  dp_no[0][v]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, -1);
  cout << ll(ans / ll(k + k)) + ll(ans2 / ll(k)) << endl;
}
