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 char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return (p1 == p2) &&
                 (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  int f = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) f ^= (ch == '-');
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
  return f ? -x : x;
}
template <class T>
inline void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 | 48);
}
template <class T>
inline void wrn(T x) {
  write(x), putchar(' ');
}
template <class T>
inline void wln(T x) {
  write(x), putchar('\n');
}
template <class T>
inline void wlnn(T x, T y) {
  wrn(x), wln(y);
}
const int N = 2e5 + 5;
struct Edge {
  int To, Next;
} e[N << 1];
int cnt = 0, head[N];
inline void add(int u, int v) {
  e[++cnt] = (Edge){v, head[u]}, head[u] = cnt;
  e[++cnt] = (Edge){u, head[v]}, head[v] = cnt;
}
long long n, k, Ans, dep[N], sz[N], f[N], son[N][7];
inline void dfs1(int u, int fa) {
  sz[u] = 1, dep[u] = dep[fa] + 1;
  for (int i = head[u], v; i; i = e[i].Next) {
    if ((v = e[i].To) == fa) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    for (int i = 1; i < k; ++i) son[u][i] += son[v][i - 1];
    son[u][0] += son[v][k - 1] + sz[v];
  }
}
inline void dfs2(int u, int fa) {
  long long add[] = {0, 0, 0, 0, 0};
  if (fa) {
    for (int i = 1; i < k; ++i) add[i] = son[fa][i] - son[u][i - 1];
    add[0] = son[fa][0] - son[u][k - 1] - sz[u];
  }
  son[u][0] += add[k - 1] + n - sz[u];
  for (int i = 1; i < k; ++i) son[u][i] += add[i - 1];
  for (int i = head[u]; i; i = e[i].Next)
    if (e[i].To != fa) dfs2(e[i].To, u);
}
int main() {
  n = read(), k = read();
  for (int i = 1, u, v; i < n; ++i) u = read(), v = read(), add(u, v);
  dfs1(1, 0), dfs2(1, 0);
  for (int i = 1; i <= n; ++i) Ans += son[i][0];
  return wln(Ans >> 1), 0;
}
