#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 100;
long long n, k;
struct Node {
  long long y, Next;
} e[2 * N];
long long len, linkk[N];
long long f[N][10];
void Insert(long long x, long long y) {
  e[++len] = (Node){y, linkk[x]};
  linkk[x] = len;
}
long long siz[N];
void dfs(long long x, long long faa) {
  siz[x] = 1;
  for (long long i = linkk[x]; i; i = e[i].Next) {
    long long y = e[i].y;
    if (y == faa) continue;
    dfs(y, x);
    siz[x] += siz[y];
    f[x][0] += f[y][k - 1] + siz[y];
    for (long long j = 1; j < k; j++) f[x][j] = f[x][j] + f[y][j - 1];
  }
}
void Dfs(long long x, long long faa) {
  long long g[5] = {};
  if (faa) {
    g[0] = f[faa][0] - f[x][k - 1] - siz[x];
    for (long long i = 1; i <= k; i++) g[i] = f[faa][i] - f[x][i - 1];
    f[x][0] += g[k - 1] + n - siz[x];
    for (long long i = 1; i <= k; i++) f[x][i] += g[i - 1];
  }
  for (long long i = linkk[x]; i; i = e[i].Next)
    if (e[i].y != faa) Dfs(e[i].y, x);
}
signed main() {
  scanf("%lld %lld", &n, &k);
  for (long long i = 1, x, y; i < n; i++)
    scanf("%lld %lld", &x, &y), Insert(x, y), Insert(y, x);
  dfs(1, 0);
  Dfs(1, 0);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans += f[i][0];
  cout << ans / 2;
  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 long long N = 2e5 + 100;
long long n, k;
struct Node {
  long long y, Next;
} e[2 * N];
long long len, linkk[N];
long long f[N][10];
void Insert(long long x, long long y) {
  e[++len] = (Node){y, linkk[x]};
  linkk[x] = len;
}
long long siz[N];
void dfs(long long x, long long faa) {
  siz[x] = 1;
  for (long long i = linkk[x]; i; i = e[i].Next) {
    long long y = e[i].y;
    if (y == faa) continue;
    dfs(y, x);
    siz[x] += siz[y];
    f[x][0] += f[y][k - 1] + siz[y];
    for (long long j = 1; j < k; j++) f[x][j] = f[x][j] + f[y][j - 1];
  }
}
void Dfs(long long x, long long faa) {
  long long g[5] = {};
  if (faa) {
    g[0] = f[faa][0] - f[x][k - 1] - siz[x];
    for (long long i = 1; i <= k; i++) g[i] = f[faa][i] - f[x][i - 1];
    f[x][0] += g[k - 1] + n - siz[x];
    for (long long i = 1; i <= k; i++) f[x][i] += g[i - 1];
  }
  for (long long i = linkk[x]; i; i = e[i].Next)
    if (e[i].y != faa) Dfs(e[i].y, x);
}
signed main() {
  scanf("%lld %lld", &n, &k);
  for (long long i = 1, x, y; i < n; i++)
    scanf("%lld %lld", &x, &y), Insert(x, y), Insert(y, x);
  dfs(1, 0);
  Dfs(1, 0);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans += f[i][0];
  cout << ans / 2;
  return 0;
}

```