#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, p, cnt;
long long dp[maxn][6], sz[maxn][6], ans = 0;
struct edge {
  int v;
  edge *next;
} e[maxn << 1], *head[maxn];
void dfs(int u, int fa) {
  sz[u][0] = 1;
  for (edge *k = head[u]; k; k = k->next)
    if (k->v != fa) {
      dfs(k->v, u);
      for (int i = p; i >= 1; i--)
        dp[u][i] += dp[k->v][i - 1], sz[u][i] += sz[k->v][i - 1];
      sz[u][0] += sz[k->v][0], dp[u][0] += sz[k->v][0] + dp[k->v][p - 1];
    }
}
void Dp(int u, int fa) {
  ans += dp[u][0];
  for (edge *k = head[u]; k; k = k->next)
    if (k->v != fa) {
      for (int i = 1; i <= p; i++)
        dp[u][i] -= dp[k->v][i - 1], sz[u][i] -= sz[k->v][i - 1];
      sz[u][0] -= sz[k->v][0], dp[u][0] -= sz[k->v][0] + dp[k->v][p - 1];
      for (int i = 1; i <= p; i++)
        dp[k->v][i] += dp[u][i - 1], sz[k->v][i] += sz[u][i - 1];
      sz[k->v][0] += sz[u][0], dp[k->v][0] += sz[u][0] + dp[u][p - 1];
      Dp(k->v, u);
      for (int i = 1; i <= p; i++)
        dp[k->v][i] -= dp[u][i - 1], sz[k->v][i] -= sz[u][i - 1];
      sz[k->v][0] -= sz[u][0], dp[k->v][0] -= sz[u][0] + dp[u][p - 1];
      for (int i = p; i >= 1; i--)
        dp[u][i] += dp[k->v][i - 1], sz[u][i] += sz[k->v][i - 1];
      sz[u][0] += sz[k->v][0], dp[u][0] += sz[k->v][0] + dp[k->v][p - 1];
    }
}
int main() {
  scanf("%d%d", &n, &p);
  int u, v;
  for (int i = 1; i < n; i++)
    scanf("%d%d", &u, &v), (e[cnt] = (edge){v, head[u]}, head[u] = &e[cnt++]),
        (e[cnt] = (edge){u, head[v]}, head[v] = &e[cnt++]);
  dfs(1, 0);
  Dp(1, 0);
  printf("%I64d", ans / 2);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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;
const int maxn = 2e5 + 10;
int n, p, cnt;
long long dp[maxn][6], sz[maxn][6], ans = 0;
struct edge {
  int v;
  edge *next;
} e[maxn << 1], *head[maxn];
void dfs(int u, int fa) {
  sz[u][0] = 1;
  for (edge *k = head[u]; k; k = k->next)
    if (k->v != fa) {
      dfs(k->v, u);
      for (int i = p; i >= 1; i--)
        dp[u][i] += dp[k->v][i - 1], sz[u][i] += sz[k->v][i - 1];
      sz[u][0] += sz[k->v][0], dp[u][0] += sz[k->v][0] + dp[k->v][p - 1];
    }
}
void Dp(int u, int fa) {
  ans += dp[u][0];
  for (edge *k = head[u]; k; k = k->next)
    if (k->v != fa) {
      for (int i = 1; i <= p; i++)
        dp[u][i] -= dp[k->v][i - 1], sz[u][i] -= sz[k->v][i - 1];
      sz[u][0] -= sz[k->v][0], dp[u][0] -= sz[k->v][0] + dp[k->v][p - 1];
      for (int i = 1; i <= p; i++)
        dp[k->v][i] += dp[u][i - 1], sz[k->v][i] += sz[u][i - 1];
      sz[k->v][0] += sz[u][0], dp[k->v][0] += sz[u][0] + dp[u][p - 1];
      Dp(k->v, u);
      for (int i = 1; i <= p; i++)
        dp[k->v][i] -= dp[u][i - 1], sz[k->v][i] -= sz[u][i - 1];
      sz[k->v][0] -= sz[u][0], dp[k->v][0] -= sz[u][0] + dp[u][p - 1];
      for (int i = p; i >= 1; i--)
        dp[u][i] += dp[k->v][i - 1], sz[u][i] += sz[k->v][i - 1];
      sz[u][0] += sz[k->v][0], dp[u][0] += sz[k->v][0] + dp[k->v][p - 1];
    }
}
int main() {
  scanf("%d%d", &n, &p);
  int u, v;
  for (int i = 1; i < n; i++)
    scanf("%d%d", &u, &v), (e[cnt] = (edge){v, head[u]}, head[u] = &e[cnt++]),
        (e[cnt] = (edge){u, head[v]}, head[v] = &e[cnt++]);
  dfs(1, 0);
  Dp(1, 0);
  printf("%I64d", ans / 2);
  return 0;
}

```