#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
  return f ? s : -s;
}
const long long N = 2e5 + 10;
long long n, k;
long long tot, head[N], ver[N << 1], nxt[N << 1];
void add(long long x, long long y) {
  nxt[++tot] = head[x];
  ver[tot] = y;
  head[x] = tot;
}
long long f[N][7], Sz[N];
void dfs_first(long long x, long long _fa) {
  Sz[x] = 1;
  for (register long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == _fa) continue;
    dfs_first(y, x);
    Sz[x] += Sz[y];
    f[x][0] += f[y][k - 1] + Sz[y];
    for (register long long j = 1; j < k; ++j) f[x][j] += f[y][j - 1];
  }
}
long long g[7];
void dfs_second(long long x, long long _fa) {
  for (register long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == _fa) continue;
    memset(g, 0, sizeof g);
    g[0] = f[x][0] - f[y][k - 1] - Sz[y];
    for (register long long j = 1; j < k; ++j) g[j] = f[x][j] - f[y][j - 1];
    f[y][0] += g[k - 1] + n - Sz[y];
    for (register long long j = 1; j < k; ++j) f[y][j] += g[j - 1];
    dfs_second(y, x);
  }
}
signed main() {
  n = read();
  k = read();
  for (register long long i = 1; i < n; ++i) {
    long long x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs_first(1, 0);
  dfs_second(1, 0);
  long long ans = 0;
  for (register long long i = 1; i <= n; ++i) ans += f[i][0];
  printf("%lld\n", ans / 2);
  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;
inline long long read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
  return f ? s : -s;
}
const long long N = 2e5 + 10;
long long n, k;
long long tot, head[N], ver[N << 1], nxt[N << 1];
void add(long long x, long long y) {
  nxt[++tot] = head[x];
  ver[tot] = y;
  head[x] = tot;
}
long long f[N][7], Sz[N];
void dfs_first(long long x, long long _fa) {
  Sz[x] = 1;
  for (register long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == _fa) continue;
    dfs_first(y, x);
    Sz[x] += Sz[y];
    f[x][0] += f[y][k - 1] + Sz[y];
    for (register long long j = 1; j < k; ++j) f[x][j] += f[y][j - 1];
  }
}
long long g[7];
void dfs_second(long long x, long long _fa) {
  for (register long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == _fa) continue;
    memset(g, 0, sizeof g);
    g[0] = f[x][0] - f[y][k - 1] - Sz[y];
    for (register long long j = 1; j < k; ++j) g[j] = f[x][j] - f[y][j - 1];
    f[y][0] += g[k - 1] + n - Sz[y];
    for (register long long j = 1; j < k; ++j) f[y][j] += g[j - 1];
    dfs_second(y, x);
  }
}
signed main() {
  n = read();
  k = read();
  for (register long long i = 1; i < n; ++i) {
    long long x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs_first(1, 0);
  dfs_second(1, 0);
  long long ans = 0;
  for (register long long i = 1; i <= n; ++i) ans += f[i][0];
  printf("%lld\n", ans / 2);
  return 0;
}

```