#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, k, ans;
long long head[200020], pos;
struct edge {
  long long to, next, c;
} e[200020 << 1];
void add(long long a, long long b, long long c) {
  pos++;
  e[pos].to = b, e[pos].c = c, e[pos].next = head[a], head[a] = pos;
}
void insert(long long a, long long b, long long c) {
  add(a, b, c);
  add(b, a, c);
}
long long size[200020], f[200020], h[200020], g[200020][5];
void dp(long long u, long long fa) {
  f[u] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    long long t = e[i].c % k, s = e[i].c / k;
    if (v == fa) continue;
    dp(v, u);
    h[v] = f[v] + g[v][0];
    ans += f[u] * size[v] + size[u] * h[v];
    size[u] += size[v];
    f[u] += f[v] + g[v][0];
    for (long long j = 0; j < k; j++)
      for (long long t = 0; t < k; t++)
        if ((t + j) < k && j + 1 && t) ans -= g[u][t] * g[v][j];
    for (long long j = 0; j < k; j++) {
      long long to = (t + j) % k;
      g[u][to] += g[v][j];
    }
  }
  ans += f[u];
  g[u][0]++;
  size[u]++;
}
int main() {
  n = read(), k = read();
  for (long long i = 1; i < n; i++) {
    long long x = read(), y = read();
    insert(x, y, 1);
  }
  dp(1, 0);
  printf("%lld\n", ans);
}

### Prompt

Please create a solution in CPP 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 x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, k, ans;
long long head[200020], pos;
struct edge {
  long long to, next, c;
} e[200020 << 1];
void add(long long a, long long b, long long c) {
  pos++;
  e[pos].to = b, e[pos].c = c, e[pos].next = head[a], head[a] = pos;
}
void insert(long long a, long long b, long long c) {
  add(a, b, c);
  add(b, a, c);
}
long long size[200020], f[200020], h[200020], g[200020][5];
void dp(long long u, long long fa) {
  f[u] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    long long t = e[i].c % k, s = e[i].c / k;
    if (v == fa) continue;
    dp(v, u);
    h[v] = f[v] + g[v][0];
    ans += f[u] * size[v] + size[u] * h[v];
    size[u] += size[v];
    f[u] += f[v] + g[v][0];
    for (long long j = 0; j < k; j++)
      for (long long t = 0; t < k; t++)
        if ((t + j) < k && j + 1 && t) ans -= g[u][t] * g[v][j];
    for (long long j = 0; j < k; j++) {
      long long to = (t + j) % k;
      g[u][to] += g[v][j];
    }
  }
  ans += f[u];
  g[u][0]++;
  size[u]++;
}
int main() {
  n = read(), k = read();
  for (long long i = 1; i < n; i++) {
    long long x = read(), y = read();
    insert(x, y, 1);
  }
  dp(1, 0);
  printf("%lld\n", ans);
}

```