#include <bits/stdc++.h>
using namespace std;
const int maxN = 200010;
struct pth {
  pth *nx;
  int nd;
} E[maxN * 2], *fir[maxN], *e;
int N, K;
int f[maxN][7], q[maxN], fa[maxN], sz[maxN], dep[maxN];
void add_edge(int x, int y) {
  e->nx = fir[x];
  e->nd = y;
  fir[x] = e++;
}
int main() {
  scanf("%d%d", &N, &K);
  e = E;
  for (int i = 1; i < N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    add_edge(x, y);
    add_edge(y, x);
  }
  q[0] = 0;
  memset(fa, -1, sizeof fa);
  for (int h = 0, t = 1; h < t; h++) {
    int p = q[h];
    for (pth *u = fir[p]; u; u = u->nx) {
      if (u->nd != fa[p]) {
        q[t++] = u->nd;
        fa[u->nd] = p;
        dep[u->nd] = dep[p] + 1;
      }
    }
  }
  for (int i = N - 1; i > -1; i--) {
    int p = q[i];
    sz[p]++;
    if (fa[p] != -1) sz[fa[p]] += sz[p];
  }
  long long ans = 0;
  for (int i = N - 1; i > -1; i--) {
    int p = q[i];
    for (pth *u = fir[p]; u; u = u->nx) {
      if (u->nd != fa[p]) {
        for (int s = 0; s < K; s++) {
          for (int t = 0; t < K; t++) {
            if (s + t) {
              ans += 1LL * f[p][s] * f[u->nd][t];
              if (s + t > K) ans += 1LL * f[p][s] * f[u->nd][t];
            }
          }
        }
        for (int s = 0; s < K; s++) {
          f[p][s] += f[u->nd][s];
          if (s == 0) {
            ans += 1LL * f[u->nd][s] * (N - sz[u->nd]);
          } else {
            ans += f[u->nd][s];
          }
        }
      }
    }
    f[p][0]++;
    rotate(f[p], f[p] + K - 1, f[p] + K);
  }
  cout << ans << endl;
  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 = 200010;
struct pth {
  pth *nx;
  int nd;
} E[maxN * 2], *fir[maxN], *e;
int N, K;
int f[maxN][7], q[maxN], fa[maxN], sz[maxN], dep[maxN];
void add_edge(int x, int y) {
  e->nx = fir[x];
  e->nd = y;
  fir[x] = e++;
}
int main() {
  scanf("%d%d", &N, &K);
  e = E;
  for (int i = 1; i < N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    add_edge(x, y);
    add_edge(y, x);
  }
  q[0] = 0;
  memset(fa, -1, sizeof fa);
  for (int h = 0, t = 1; h < t; h++) {
    int p = q[h];
    for (pth *u = fir[p]; u; u = u->nx) {
      if (u->nd != fa[p]) {
        q[t++] = u->nd;
        fa[u->nd] = p;
        dep[u->nd] = dep[p] + 1;
      }
    }
  }
  for (int i = N - 1; i > -1; i--) {
    int p = q[i];
    sz[p]++;
    if (fa[p] != -1) sz[fa[p]] += sz[p];
  }
  long long ans = 0;
  for (int i = N - 1; i > -1; i--) {
    int p = q[i];
    for (pth *u = fir[p]; u; u = u->nx) {
      if (u->nd != fa[p]) {
        for (int s = 0; s < K; s++) {
          for (int t = 0; t < K; t++) {
            if (s + t) {
              ans += 1LL * f[p][s] * f[u->nd][t];
              if (s + t > K) ans += 1LL * f[p][s] * f[u->nd][t];
            }
          }
        }
        for (int s = 0; s < K; s++) {
          f[p][s] += f[u->nd][s];
          if (s == 0) {
            ans += 1LL * f[u->nd][s] * (N - sz[u->nd]);
          } else {
            ans += f[u->nd][s];
          }
        }
      }
    }
    f[p][0]++;
    rotate(f[p], f[p] + K - 1, f[p] + K);
  }
  cout << ans << endl;
  return 0;
}

```