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 = 3e5 + 10;
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) - '0' + ch;
    ch = getchar();
  }
  return x * f;
}
int n, k, size[N];
long long ans, f[N][5];
vector<int> g[N];
void dfs(int u, int fa, int d) {
  size[u] = 1;
  f[u][d % k] = 1;
  for (auto &v : g[u]) {
    if (v == fa) continue;
    dfs(v, u, d + 1);
    size[u] += size[v];
    ans += 1ll * (n - size[v]) * size[v];
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        int z = ((i + j - 2 * d) % k + k) % k;
        z = (k - z) % k;
        ans += 1ll * z * f[u][i] * f[v][j];
      }
    }
    for (int i = 0; i < k; i++) f[u][i] += f[v][i];
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n - 1; i++) {
    int a = read(), b = read();
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, 0, 0);
  printf("%lld\n", ans / k);
}
