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;
inline long long max(long long x, long long y) { return x > y ? x : y; }
inline long long min(long long x, long long y) { return x > y ? y : x; }
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long N = 200000 + 10;
long long n, k, f[N], g[10][N], ans = 0, tmp[10], cnt, head[N];
struct node {
  long long to, next;
} edge[N << 1];
inline void add(long long u, long long v) {
  edge[++cnt] = {v, head[u]};
  head[u] = cnt;
}
inline void dfs(long long u, long long fa) {
  g[0][u] = 1;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    f[u] += f[v] + g[0][v];
    for (long long i = 0; i < k; i++) g[(i + 1) % k][u] += g[i][v];
  }
}
inline void solve(long long u, long long fa) {
  ans += f[u];
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    for (long long i = 0; i < k; i++) tmp[i] = g[i][v];
    for (long long i = 0; i < k; i++)
      g[i][v] += g[(i - 1 + 2 * k) % k][u] - tmp[(i - 2 + 2 * k) % k];
    f[v] = f[u] - tmp[0] + g[0][u] - tmp[k - 1];
    solve(v, u);
  }
}
signed main() {
  n = read();
  k = read();
  long long u, v;
  for (long long i = 1; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  solve(1, 0);
  cout << ans / 2 << endl;
}
