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 long long N = 1e6 + 5;
long long n, m, ans, f[N][10], Ans[N];
vector<long long> g[N];
inline long long read() {
  long long ret = 0, f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - 48;
    c = getchar();
  }
  if (f) return -ret;
  return ret;
}
void dfs(long long u, long long fa) {
  f[u][0]++;
  for (long long i = 0; i < g[u].size(); i++) {
    long long v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    Ans[u] = Ans[u] + Ans[v] + f[v][0];
    for (long long j = 0; j < m; j++) f[u][(j + 1) % m] += f[v][j];
  }
}
void solve(long long u, long long fa) {
  ans += Ans[u];
  for (long long i = 0; i < g[u].size(); i++) {
    long long v = g[u][i];
    if (v == fa) continue;
    Ans[u] = Ans[u] - Ans[v] - f[v][0];
    for (long long j = 0; j < m; j++) f[u][(j + 1) % m] -= f[v][j];
    Ans[v] = Ans[v] + Ans[u] + f[u][0];
    for (long long j = 0; j < m; j++) f[v][(j + 1) % m] += f[u][j];
    solve(v, u);
    for (long long j = 0; j < m; j++) f[v][(j + 1) % m] -= f[u][j];
    Ans[v] = Ans[v] - Ans[u] - f[u][0];
    for (long long j = 0; j < m; j++) f[u][(j + 1) % m] += f[v][j];
    Ans[u] = Ans[u] + Ans[v] + f[v][0];
  }
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i < n; i++) {
    long long x = read(), y = read();
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  solve(1, 0);
  cout << ans / 2;
  return 0;
}
