#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 30;
const int MOD = 1e9 + 7;
const int N = 2e5 + 10;
vector<int> a[N];
long long sum[N][5], cnt[N][5];
int m;
long long ret;
void DFS(int u, int parent) {
  cnt[u][0] = 1;
  for (auto& v : a[u]) {
    if (v == parent) continue;
    DFS(v, u);
    long long cur = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < m; ++j) {
        int extra = (i + j + 1 + m - 1) / m;
        cur += cnt[u][i] * cnt[v][j] * extra;
        cur += cnt[u][i] * sum[v][j];
        cur += cnt[v][j] * sum[u][i];
      }
    }
    ret += cur;
    for (int i = 0; i < m; ++i) {
      int j = (i + 1) % m;
      cnt[u][j] += cnt[v][i];
      sum[u][j] += sum[v][i];
      if (j == 0) sum[u][j] += cnt[v][i];
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x;
    --y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  ret = 0;
  DFS(0, -1);
  printf("%lld\n", ret);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 30;
const int MOD = 1e9 + 7;
const int N = 2e5 + 10;
vector<int> a[N];
long long sum[N][5], cnt[N][5];
int m;
long long ret;
void DFS(int u, int parent) {
  cnt[u][0] = 1;
  for (auto& v : a[u]) {
    if (v == parent) continue;
    DFS(v, u);
    long long cur = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < m; ++j) {
        int extra = (i + j + 1 + m - 1) / m;
        cur += cnt[u][i] * cnt[v][j] * extra;
        cur += cnt[u][i] * sum[v][j];
        cur += cnt[v][j] * sum[u][i];
      }
    }
    ret += cur;
    for (int i = 0; i < m; ++i) {
      int j = (i + 1) % m;
      cnt[u][j] += cnt[v][i];
      sum[u][j] += sum[v][i];
      if (j == 0) sum[u][j] += cnt[v][i];
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x;
    --y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  ret = 0;
  DFS(0, -1);
  printf("%lld\n", ret);
  return 0;
}

```