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;
struct ed {
  int u, nex;
} e[200010 * 2];
long long n, k0, fir[200010], st;
long long ans, jp[200010], s[200010][5], Jp[200010], S[200010][5];
void add(long long u, long long v) {
  e[++st].u = v;
  e[st].nex = fir[u];
  fir[u] = st;
  e[++st].u = u;
  e[st].nex = fir[v];
  fir[v] = st;
}
void dfs1(long long u, long long fa) {
  s[u][0] = 1;
  jp[u] = 0;
  for (long long i = fir[u]; i; i = e[i].nex) {
    long long v = e[i].u;
    if (v == fa) continue;
    dfs1(v, u);
    jp[u] += jp[v];
    for (long long k = 0; k < k0; k++) s[u][k] += s[v][(k - 1 + k0) % k0];
    jp[u] += s[v][0];
  }
}
void dfs2(long long u, long long fa) {
  if (u != 1) {
    for (long long k = 0; k < k0; k++) {
      S[u][k] = S[fa][(k - 1 + k0) % k0];
      S[u][k] = S[u][k] - s[u][(k - 2 + k0) % k0] + s[u][k];
    }
    Jp[u] = Jp[fa];
    Jp[u] += (S[fa][0] - s[u][k0 - 1]);
    Jp[u] -= s[u][0];
  }
  ans += Jp[u];
  for (long long i = fir[u]; i; i = e[i].nex) {
    long long v = e[i].u;
    if (v == fa) continue;
    dfs2(v, u);
  }
}
int main() {
  cin >> n >> k0;
  for (int i = 1, u, v; i < n; i++) cin >> u >> v, add(u, v);
  dfs1(1, 0);
  memcpy(S, s, sizeof(s));
  Jp[1] = jp[1];
  dfs2(1, 0);
  cout << ans / 2 << endl;
}
