#include <bits/stdc++.h>
using namespace std;
static struct { int v, next; } edge[2 * 200010];
static int N, K, G, head[200010];
static long long ans, dp[200010][10], sum[10];
inline void add_edge(int u, int v) {
  edge[G].v = v;
  edge[G].next = head[u];
  head[u] = G++;
  edge[G].v = u;
  edge[G].next = head[v];
  head[v] = G++;
}
int dfs(int x, int fa) {
  int size = 1, tmp;
  dp[x][0] = 1;
  for (int i = head[x]; ~i; i = edge[i].next)
    if (edge[i].v != fa) {
      size += (tmp = dfs(edge[i].v, x));
      ans += (long long)tmp * (N - tmp);
      for (int j = 0; j < K; j++)
        for (int k = 0; k < K; k++)
          sum[(j + k + 1) % K] += dp[x][j] * dp[edge[i].v][k];
      for (int j = 0; j < K; j++) dp[x][j] += dp[edge[i].v][(K + j - 1) % K];
    }
  return size;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &N, &K);
  for (int i = 1, u, v; i < N; i++) scanf("%d%d", &u, &v), add_edge(u, v);
  dfs(1, 0);
  for (int i = 1; i < K; i++) ans += sum[i] * (K - i);
  printf("%lld\n", ans / K);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
static struct { int v, next; } edge[2 * 200010];
static int N, K, G, head[200010];
static long long ans, dp[200010][10], sum[10];
inline void add_edge(int u, int v) {
  edge[G].v = v;
  edge[G].next = head[u];
  head[u] = G++;
  edge[G].v = u;
  edge[G].next = head[v];
  head[v] = G++;
}
int dfs(int x, int fa) {
  int size = 1, tmp;
  dp[x][0] = 1;
  for (int i = head[x]; ~i; i = edge[i].next)
    if (edge[i].v != fa) {
      size += (tmp = dfs(edge[i].v, x));
      ans += (long long)tmp * (N - tmp);
      for (int j = 0; j < K; j++)
        for (int k = 0; k < K; k++)
          sum[(j + k + 1) % K] += dp[x][j] * dp[edge[i].v][k];
      for (int j = 0; j < K; j++) dp[x][j] += dp[edge[i].v][(K + j - 1) % K];
    }
  return size;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &N, &K);
  for (int i = 1, u, v; i < N; i++) scanf("%d%d", &u, &v), add_edge(u, v);
  dfs(1, 0);
  for (int i = 1; i < K; i++) ans += sum[i] * (K - i);
  printf("%lld\n", ans / K);
  return 0;
}

```