#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e6 + 10;
struct edge {
  long long nt, to;
} E[MAXN];
long long N, head[MAXN], cnt = -1, ans = 0, K, f[MAXN][7], siz[MAXN];
void add(long long, long long);
void dfs(long long, long long);
void dfs2(long long, long long);
int main() {
  memset(head, -1, sizeof(head));
  scanf("%lld%lld", &N, &K);
  for (long long i = 1, x, y; i < N; i++) {
    scanf("%lld%lld", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans / 2);
  return 0;
}
void dfs2(long long n, long long ff) {
  long long t[7] = {0};
  if (ff) {
    for (long long i = 1; i < K; i++) {
      t[i] = f[ff][i] - f[n][i - 1];
    }
    t[0] = f[ff][0] - f[n][K - 1] - siz[n];
    for (long long i = 1; i < K; i++) {
      f[n][i] += t[i - 1];
    }
    f[n][0] += t[K - 1] + (N - siz[n]);
  }
  ans += f[n][0];
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff)
      continue;
    else {
      dfs2(v, n);
    }
  }
}
void dfs(long long n, long long ff) {
  siz[n] = 1;
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff)
      continue;
    else {
      dfs(v, n);
      siz[n] += siz[v];
      f[n][0] += f[v][K - 1] + siz[v];
      for (long long i = 1; i < K; i++) {
        f[n][i] += f[v][i - 1];
      }
    }
  }
}
void add(long long x, long long y) {
  cnt++;
  E[cnt].to = y;
  E[cnt].nt = head[x];
  head[x] = cnt;
}

### 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 long long MAXN = 1e6 + 10;
struct edge {
  long long nt, to;
} E[MAXN];
long long N, head[MAXN], cnt = -1, ans = 0, K, f[MAXN][7], siz[MAXN];
void add(long long, long long);
void dfs(long long, long long);
void dfs2(long long, long long);
int main() {
  memset(head, -1, sizeof(head));
  scanf("%lld%lld", &N, &K);
  for (long long i = 1, x, y; i < N; i++) {
    scanf("%lld%lld", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans / 2);
  return 0;
}
void dfs2(long long n, long long ff) {
  long long t[7] = {0};
  if (ff) {
    for (long long i = 1; i < K; i++) {
      t[i] = f[ff][i] - f[n][i - 1];
    }
    t[0] = f[ff][0] - f[n][K - 1] - siz[n];
    for (long long i = 1; i < K; i++) {
      f[n][i] += t[i - 1];
    }
    f[n][0] += t[K - 1] + (N - siz[n]);
  }
  ans += f[n][0];
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff)
      continue;
    else {
      dfs2(v, n);
    }
  }
}
void dfs(long long n, long long ff) {
  siz[n] = 1;
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff)
      continue;
    else {
      dfs(v, n);
      siz[n] += siz[v];
      f[n][0] += f[v][K - 1] + siz[v];
      for (long long i = 1; i < K; i++) {
        f[n][i] += f[v][i - 1];
      }
    }
  }
}
void add(long long x, long long y) {
  cnt++;
  E[cnt].to = y;
  E[cnt].nt = head[x];
  head[x] = cnt;
}

```