#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
long long d[maxn][5], ans;
int k, cnt;
struct node {
  int v, next;
} edge[maxn * 2 + 10];
int head[maxn], sz[maxn][5];
void add(int x, int y) {
  edge[cnt].next = head[x];
  edge[cnt].v = y;
  head[x] = cnt++;
}
void dfs(int x, int fa) {
  long long td[5], delta;
  int tsz[5], i, j, p, v;
  for (i = head[x]; i != -1; i = edge[i].next) {
    v = edge[i].v;
    if (v == fa) continue;
    dfs(v, x);
    memset(td, 0, sizeof(td));
    memset(tsz, 0, sizeof(tsz));
    for (j = 0; j < k; j++) tsz[(j + 1) % k] += sz[v][j];
    for (j = 1; j < k; j++) td[(j + 1) % k] += d[v][j];
    td[1 % k] += d[v][0] + sz[v][0];
    for (j = 0; j < k; j++) {
      for (p = 0; p < k; p++) {
        delta = (p + j + k - 1) / k - (p > 0) - (j > 0);
        ans += d[x][j] * tsz[p] + td[p] * sz[x][j];
        if (delta) ans += (long long)delta * sz[x][j] * tsz[p];
      }
    }
    for (j = 0; j < k; j++) {
      d[x][j] += td[j];
      sz[x][j] += tsz[j];
    }
  }
  for (i = 0; i < k; i++) {
    ans += d[x][i];
  }
  sz[x][0]++;
}
int main() {
  int n, x, y, i;
  scanf("%d%d", &n, &k);
  cnt = 0;
  memset(head, -1, sizeof(head));
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  ans = 0;
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 = 200010;
long long d[maxn][5], ans;
int k, cnt;
struct node {
  int v, next;
} edge[maxn * 2 + 10];
int head[maxn], sz[maxn][5];
void add(int x, int y) {
  edge[cnt].next = head[x];
  edge[cnt].v = y;
  head[x] = cnt++;
}
void dfs(int x, int fa) {
  long long td[5], delta;
  int tsz[5], i, j, p, v;
  for (i = head[x]; i != -1; i = edge[i].next) {
    v = edge[i].v;
    if (v == fa) continue;
    dfs(v, x);
    memset(td, 0, sizeof(td));
    memset(tsz, 0, sizeof(tsz));
    for (j = 0; j < k; j++) tsz[(j + 1) % k] += sz[v][j];
    for (j = 1; j < k; j++) td[(j + 1) % k] += d[v][j];
    td[1 % k] += d[v][0] + sz[v][0];
    for (j = 0; j < k; j++) {
      for (p = 0; p < k; p++) {
        delta = (p + j + k - 1) / k - (p > 0) - (j > 0);
        ans += d[x][j] * tsz[p] + td[p] * sz[x][j];
        if (delta) ans += (long long)delta * sz[x][j] * tsz[p];
      }
    }
    for (j = 0; j < k; j++) {
      d[x][j] += td[j];
      sz[x][j] += tsz[j];
    }
  }
  for (i = 0; i < k; i++) {
    ans += d[x][i];
  }
  sz[x][0]++;
}
int main() {
  int n, x, y, i;
  scanf("%d%d", &n, &k);
  cnt = 0;
  memset(head, -1, sizeof(head));
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  ans = 0;
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

```