#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long long maxn = 400005;
struct Edge {
  long long nxt, to;
} e[maxn << 1];
long long hd[maxn], e_cnt;
void add(long long u, long long v) {
  e[++e_cnt] = (Edge){hd[u], v};
  hd[u] = e_cnt;
}
long long cas, n, k, ans, f[maxn][5], siz[maxn], c[5];
void dfs1(long long u, long long r) {
  siz[u] = 1;
  for (long long i = hd[u]; i; i = e[i].nxt)
    if (e[i].to != r) {
      dfs1(e[i].to, u);
      siz[u] += siz[e[i].to];
      for (long long j = 0; j <= k - 1; j++) c[(j + 1) % k] = f[e[i].to][j];
      long long rec = 0;
      for (long long j = 1; j <= k - 1; j++) {
        rec += f[u][k - j];
        ans -= rec * c[j];
      }
      for (long long j = 0; j <= k - 1; j++) f[u][j] += c[j];
    }
}
void dfs2(long long u, long long r) {
  for (long long i = hd[u]; i; i = e[i].nxt)
    if (e[i].to != r) {
      long long tmp = f[u][k - 1];
      tmp -= f[e[i].to][max(k - 2, 0ll)];
      ans += tmp * siz[e[i].to];
      for (long long j = 0; j <= k - 1; j++)
        c[(j + 1) % k] = f[u][j] - f[e[i].to][(j - 1 + k) % k];
      for (long long j = 0; j <= k - 1; j++) f[e[i].to][j] += c[j];
      dfs2(e[i].to, u);
    }
}
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    add(a, b);
    add(b, a);
  }
  for (long long i = 1; i <= n; i++) f[i][0] = 1;
  dfs1(1, 0);
  dfs2(1, 0);
  long long tmp = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= k - 1; j++) tmp += f[i][j];
  ans += tmp >> 1;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long long maxn = 400005;
struct Edge {
  long long nxt, to;
} e[maxn << 1];
long long hd[maxn], e_cnt;
void add(long long u, long long v) {
  e[++e_cnt] = (Edge){hd[u], v};
  hd[u] = e_cnt;
}
long long cas, n, k, ans, f[maxn][5], siz[maxn], c[5];
void dfs1(long long u, long long r) {
  siz[u] = 1;
  for (long long i = hd[u]; i; i = e[i].nxt)
    if (e[i].to != r) {
      dfs1(e[i].to, u);
      siz[u] += siz[e[i].to];
      for (long long j = 0; j <= k - 1; j++) c[(j + 1) % k] = f[e[i].to][j];
      long long rec = 0;
      for (long long j = 1; j <= k - 1; j++) {
        rec += f[u][k - j];
        ans -= rec * c[j];
      }
      for (long long j = 0; j <= k - 1; j++) f[u][j] += c[j];
    }
}
void dfs2(long long u, long long r) {
  for (long long i = hd[u]; i; i = e[i].nxt)
    if (e[i].to != r) {
      long long tmp = f[u][k - 1];
      tmp -= f[e[i].to][max(k - 2, 0ll)];
      ans += tmp * siz[e[i].to];
      for (long long j = 0; j <= k - 1; j++)
        c[(j + 1) % k] = f[u][j] - f[e[i].to][(j - 1 + k) % k];
      for (long long j = 0; j <= k - 1; j++) f[e[i].to][j] += c[j];
      dfs2(e[i].to, u);
    }
}
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    add(a, b);
    add(b, a);
  }
  for (long long i = 1; i <= n; i++) f[i][0] = 1;
  dfs1(1, 0);
  dfs2(1, 0);
  long long tmp = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= k - 1; j++) tmp += f[i][j];
  ans += tmp >> 1;
  cout << ans << endl;
  return 0;
}

```