#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T poww(T b, T p) {
  long long a = 1;
  while (p) {
    if (p & 1) {
      a = (a * b);
    }
    p >>= 1;
    b = (b * b);
  }
  return a;
}
template <class T>
inline T gcd(T a, T b) {
  if (b > a) return gcd(b, a);
  return ((b == 0) ? a : gcd(b, a % b));
}
int n, k;
std::vector<std::vector<int> > edges;
long long ans = 0;
long long remainders[200010][5];
int dfs(int v, int par) {
  int i, j, l = edges[v].size();
  long long totalnode = 1;
  for ((i) = 0; (i) < (l); (i)++) {
    int c = edges[v][i];
    if (c == par) continue;
    totalnode += dfs(c, v);
    int a, b;
    std::vector<long long> temp(5, 0);
    for ((a) = 0; (a) < (k); (a)++) {
      for ((b) = 0; (b) < (k); (b)++) {
        int bb = (b + 1) % k;
        j = (bb + a) % k;
        ans += ((k - j) % k) * remainders[v][a] * remainders[c][b];
      }
    }
    for ((b) = 0; (b) < (k); (b)++) {
      j = (b + 1) % k;
      remainders[v][j] += remainders[c][b];
      ans += ((k - j) % k) * remainders[c][b];
    }
  }
  remainders[v][0]++;
  ans += totalnode * (n - totalnode);
  return totalnode;
}
int main() {
  register int i, j;
  scanf("%d%d", &n, &k);
  int a, b;
  edges.resize(n);
  for ((i) = 0; (i) < (n - 1); (i)++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(0, -1);
  printf("%lld\n", ans / k);
  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;
template <class T>
inline T poww(T b, T p) {
  long long a = 1;
  while (p) {
    if (p & 1) {
      a = (a * b);
    }
    p >>= 1;
    b = (b * b);
  }
  return a;
}
template <class T>
inline T gcd(T a, T b) {
  if (b > a) return gcd(b, a);
  return ((b == 0) ? a : gcd(b, a % b));
}
int n, k;
std::vector<std::vector<int> > edges;
long long ans = 0;
long long remainders[200010][5];
int dfs(int v, int par) {
  int i, j, l = edges[v].size();
  long long totalnode = 1;
  for ((i) = 0; (i) < (l); (i)++) {
    int c = edges[v][i];
    if (c == par) continue;
    totalnode += dfs(c, v);
    int a, b;
    std::vector<long long> temp(5, 0);
    for ((a) = 0; (a) < (k); (a)++) {
      for ((b) = 0; (b) < (k); (b)++) {
        int bb = (b + 1) % k;
        j = (bb + a) % k;
        ans += ((k - j) % k) * remainders[v][a] * remainders[c][b];
      }
    }
    for ((b) = 0; (b) < (k); (b)++) {
      j = (b + 1) % k;
      remainders[v][j] += remainders[c][b];
      ans += ((k - j) % k) * remainders[c][b];
    }
  }
  remainders[v][0]++;
  ans += totalnode * (n - totalnode);
  return totalnode;
}
int main() {
  register int i, j;
  scanf("%d%d", &n, &k);
  int a, b;
  edges.resize(n);
  for ((i) = 0; (i) < (n - 1); (i)++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(0, -1);
  printf("%lld\n", ans / k);
  return 0;
}

```