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>
struct link {
  link *nxt;
  int p;
};
const int N = 200000 + 10, K = 5 + 2;
int n, k;
long long ans, f[N][K], g[N];
link poool[N << 1], *pool = poool, *head[N];
inline void connect(int u, int v) {
  pool->p = v, pool->nxt = head[u], head[u] = pool++;
}
inline void dplink(int p, int s) {
  for (int i = 0; i < k - 1; ++i) f[p][i + 1] += f[s][i];
  g[p] += g[s] + f[s][k - 1], f[p][0] += f[s][k - 1];
}
inline void undplink(int p, int s) {
  for (int i = 0; i < k - 1; ++i) f[p][i + 1] -= f[s][i];
  g[p] -= g[s] + f[s][k - 1], f[p][0] -= f[s][k - 1];
}
inline void revlink(int p, int s) { undplink(p, s), dplink(s, p); }
void dfs1(int p, int fa) {
  f[p][0] = 1;
  for (link *e = head[p]; e; e = e->nxt)
    if (e->p != fa) dfs1(e->p, p), dplink(p, e->p);
}
void dfs2(int p, int fa) {
  for (int i = 1; i < k; ++i) ans += f[p][i];
  ans += g[p];
  for (link *e = head[p]; e; e = e->nxt)
    if (e->p != fa) revlink(p, e->p), dfs2(e->p, p), revlink(e->p, p);
}
int main(void) {
  scanf("%d %d", &n, &k);
  for (int i = 2, u, v; i <= n; ++i)
    scanf("%d %d", &u, &v), connect(u, v), connect(v, u);
  dfs1(1, 0), dfs2(1, 0), printf("%lld\n", ans >> 1);
}
