#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  int x = 0, f = 1;
  char a = getchar();
  while (a < '0' || a > '9') {
    if (a == '-') f = -1;
    a = getchar();
  }
  while (a >= '0' && a <= '9') x = x * 10 + a - '0', a = getchar();
  return x * f;
}
int n, k, cnt = 1, head[200005], d[200005];
long long ans, dp[200005][10], g[200005][10];
struct edges {
  int to, next;
} e[2 * 200005];
inline void insert() {
  int u = read(), v = read();
  d[u]++;
  d[v]++;
  e[++cnt] = (edges){v, head[u]};
  head[u] = cnt;
  e[++cnt] = (edges){u, head[v]};
  head[v] = cnt;
}
void dfs(int x, int fa) {
  g[x][0] = 1;
  long long delta = 0;
  for (int i = head[x]; i; i = e[i].next) {
    if (fa == e[i].to) continue;
    dfs(e[i].to, x);
    for (int a = 0; a < k; a++)
      for (int b = 0; b < k; b++) {
        if (a + b + 1 <= k)
          delta +=
              (dp[x][a] + g[x][a]) * g[e[i].to][b] + dp[e[i].to][b] * g[x][a];
        else
          delta += (dp[x][a] + 2 * g[x][a]) * g[e[i].to][b] +
                   dp[e[i].to][b] * g[x][a];
      }
    for (int a = 0; a < k; a++) g[x][(a + 1) % k] += g[e[i].to][a];
    for (int a = 0; a < k - 1; a++) dp[x][(a + 1) % k] += dp[e[i].to][a];
    dp[x][0] += dp[e[i].to][k - 1] + g[e[i].to][k - 1];
  }
  ans += delta;
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i < n; i++) insert();
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
inline long long read() {
  int x = 0, f = 1;
  char a = getchar();
  while (a < '0' || a > '9') {
    if (a == '-') f = -1;
    a = getchar();
  }
  while (a >= '0' && a <= '9') x = x * 10 + a - '0', a = getchar();
  return x * f;
}
int n, k, cnt = 1, head[200005], d[200005];
long long ans, dp[200005][10], g[200005][10];
struct edges {
  int to, next;
} e[2 * 200005];
inline void insert() {
  int u = read(), v = read();
  d[u]++;
  d[v]++;
  e[++cnt] = (edges){v, head[u]};
  head[u] = cnt;
  e[++cnt] = (edges){u, head[v]};
  head[v] = cnt;
}
void dfs(int x, int fa) {
  g[x][0] = 1;
  long long delta = 0;
  for (int i = head[x]; i; i = e[i].next) {
    if (fa == e[i].to) continue;
    dfs(e[i].to, x);
    for (int a = 0; a < k; a++)
      for (int b = 0; b < k; b++) {
        if (a + b + 1 <= k)
          delta +=
              (dp[x][a] + g[x][a]) * g[e[i].to][b] + dp[e[i].to][b] * g[x][a];
        else
          delta += (dp[x][a] + 2 * g[x][a]) * g[e[i].to][b] +
                   dp[e[i].to][b] * g[x][a];
      }
    for (int a = 0; a < k; a++) g[x][(a + 1) % k] += g[e[i].to][a];
    for (int a = 0; a < k - 1; a++) dp[x][(a + 1) % k] += dp[e[i].to][a];
    dp[x][0] += dp[e[i].to][k - 1] + g[e[i].to][k - 1];
  }
  ans += delta;
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i < n; i++) insert();
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```