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 int read() {
  int sign = 1, sum = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') sign = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    sum = (sum << 1) + (sum << 3) + c - '0';
    c = getchar();
  }
  return sign * sum;
}
const int N = 2e5 + 50;
struct Edge {
  int nxt, to;
} edge[N << 1];
int head[N], tot = 0;
inline void add(int u, int v) {
  ++tot;
  edge[tot].to = v;
  edge[tot].nxt = head[u];
  head[u] = tot;
}
inline void addedge(int u, int v) { add(v, u), add(u, v); }
int n, k;
int size[N];
long long f[N][5];
inline void dfs1(int u, int fa) {
  size[u] = 1;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].to;
    if (v != fa) {
      dfs1(v, u);
      size[u] += size[v];
      f[u][0] += f[v][k - 1] + size[v];
      for (int j = 1; j < k; ++j) f[u][j] += f[v][j - 1];
    }
  }
}
inline void dfs2(int u, int fa) {
  long long cnt[5] = {0};
  if (fa) {
    cnt[0] = f[fa][0] - f[u][k - 1] - size[u];
    for (int j = 1; j < k; ++j) cnt[j] = f[fa][j] - f[u][j - 1];
  }
  f[u][0] += cnt[k - 1] + n - size[u];
  for (int j = 1; j < k; ++j) f[u][j] += cnt[j - 1];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].to;
    if (v != fa) dfs2(v, u);
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    addedge(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += f[i][0];
  printf("%lld\n", (ans >> 1));
}
