#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
int n, k, cnt;
int head[N], fa[N], size[N];
long long p[N][6], tot[N][6], f[N][5], ftot[N][5], ans;
struct xint {
  int to, next;
} e[N * 2];
void add(int x, int y) {
  e[++cnt] = (xint){y, head[x]};
  head[x] = cnt;
  e[++cnt] = (xint){x, head[y]};
  head[y] = cnt;
}
void dfs1(int x, int d) {
  p[1][d % k] += d;
  tot[1][d % k]++;
  size[x] = 1;
  f[x][0] = 0;
  ftot[x][0] = 1;
  for (int i = head[x], y = e[i].to; i; i = e[i].next, y = e[i].to)
    if (y != fa[x]) {
      fa[y] = x;
      dfs1(y, d + 1);
      size[x] += size[y];
      for (int j = 0; j <= k - 1; ++j) {
        ftot[x][(j + 1) % k] += ftot[y][j];
        f[x][(j + 1) % k] += f[y][j] + ftot[y][j];
      }
    }
}
void dfs2(int x) {
  for (int i = 0; i <= k - 1; ++i) {
    ans += (p[x][i] - tot[x][i] * i) / k;
    if (i > 0) ans += tot[x][i];
  }
  for (int i = head[x], y = e[i].to; i; i = e[i].next, y = e[i].to)
    if (y != fa[x]) {
      for (int j = 0; j <= k - 1; ++j) {
        p[y][(j + 1) % k] = p[x][(j + 1) % k] - f[y][j] - ftot[y][j];
        tot[y][(j + 1) % k] = tot[x][(j + 1) % k] - ftot[y][j];
      }
      for (int j = k - 1; j >= 0; --j) {
        p[y][j + 1] = p[y][j] + (tot[x][j] - ftot[y][(j + k - 1) % k]);
        tot[y][j + 1] = tot[y][j];
      }
      p[y][0] = p[y][k];
      tot[y][0] = tot[y][k];
      for (int j = 0; j <= k - 1; ++j) {
        p[y][j] += f[y][j];
        tot[y][j] += ftot[y][j];
      }
      dfs2(y);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 2; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
  }
  dfs1(1, 0);
  dfs2(1);
  cout << ans / 2 << endl;
}

### Prompt

Please formulate 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 N = 200001;
int n, k, cnt;
int head[N], fa[N], size[N];
long long p[N][6], tot[N][6], f[N][5], ftot[N][5], ans;
struct xint {
  int to, next;
} e[N * 2];
void add(int x, int y) {
  e[++cnt] = (xint){y, head[x]};
  head[x] = cnt;
  e[++cnt] = (xint){x, head[y]};
  head[y] = cnt;
}
void dfs1(int x, int d) {
  p[1][d % k] += d;
  tot[1][d % k]++;
  size[x] = 1;
  f[x][0] = 0;
  ftot[x][0] = 1;
  for (int i = head[x], y = e[i].to; i; i = e[i].next, y = e[i].to)
    if (y != fa[x]) {
      fa[y] = x;
      dfs1(y, d + 1);
      size[x] += size[y];
      for (int j = 0; j <= k - 1; ++j) {
        ftot[x][(j + 1) % k] += ftot[y][j];
        f[x][(j + 1) % k] += f[y][j] + ftot[y][j];
      }
    }
}
void dfs2(int x) {
  for (int i = 0; i <= k - 1; ++i) {
    ans += (p[x][i] - tot[x][i] * i) / k;
    if (i > 0) ans += tot[x][i];
  }
  for (int i = head[x], y = e[i].to; i; i = e[i].next, y = e[i].to)
    if (y != fa[x]) {
      for (int j = 0; j <= k - 1; ++j) {
        p[y][(j + 1) % k] = p[x][(j + 1) % k] - f[y][j] - ftot[y][j];
        tot[y][(j + 1) % k] = tot[x][(j + 1) % k] - ftot[y][j];
      }
      for (int j = k - 1; j >= 0; --j) {
        p[y][j + 1] = p[y][j] + (tot[x][j] - ftot[y][(j + k - 1) % k]);
        tot[y][j + 1] = tot[y][j];
      }
      p[y][0] = p[y][k];
      tot[y][0] = tot[y][k];
      for (int j = 0; j <= k - 1; ++j) {
        p[y][j] += f[y][j];
        tot[y][j] += ftot[y][j];
      }
      dfs2(y);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 2; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
  }
  dfs1(1, 0);
  dfs2(1);
  cout << ans / 2 << endl;
}

```