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;
const int N = 200000 + 7;
const int M = 20;
const int inf = 1e9 + 7;
const long long linf = 1e18 + 7;
const double pi = acos(-1);
const double eps = 1e-7;
const bool multipleTest = 0;
long long f[N][6];
long long sum[N][6];
long long ans;
int n, k;
vector<int> adj[N];
long long g[6];
long long sl[6];
void dfs(int u, int r) {
  for (auto& v : adj[u]) {
    if (v != r) {
      dfs(v, u);
      for (int j = (1); j < (k); j++) {
        ans += sum[v][j];
        g[(j + 1) % k] = sum[v][j];
        sl[(j + 1) % k] = f[v][j];
      }
      g[1 % k] = sum[v][0] + f[v][0];
      sl[1 % k] = f[v][0];
      ans += g[1 % k];
      for (int j = (0); j < (k); j++) {
        for (int t = (0); t < (k); t++) {
          if (j == 0 || t == 0 || j + t > k) {
            ans += sum[u][j] * sl[t];
            ans += g[t] * f[u][j];
          } else {
            ans += sum[u][j] * sl[t];
            ans += g[t] * f[u][j];
            ans -= sl[t] * f[u][j];
          }
        }
      }
      for (int j = (0); j < (k); j++) {
        sum[u][j] += g[j];
        f[u][j] += sl[j];
      }
    }
  }
  f[u][0]++;
}
void solve() {
  cin >> n >> k;
  for (int i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  cout << ans << '\n';
}
int main() {
  int Test = 1;
  if (multipleTest) {
    cin >> Test;
  }
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}
