#include <bits/stdc++.h>
using namespace std;
template <class _T>
inline void rd(_T &_a) {
  int _f = 0, _ch = getchar();
  _a = 0;
  while (_ch < '0' || _ch > '9') {
    if (_ch == '-') _f = 1;
    _ch = getchar();
  }
  while (_ch >= '0' && _ch <= '9') _a = _a * 10 + _ch - '0', _ch = getchar();
  if (_f) _a = -_a;
}
template <class _T>
inline _T dmin(const _T _a, const _T _b) {
  return _a < _b ? _a : _b;
}
template <class _T>
inline _T dmax(const _T _a, const _T _b) {
  return _a > _b ? _a : _b;
}
template <class _T>
inline bool cmin(_T &_a, const _T _b) {
  return _a > _b ? _a = _b, 1 : 0;
}
template <class _T>
inline bool cmax(_T &_a, const _T _b) {
  return _a < _b ? _a = _b, 1 : 0;
}
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int N = 2e5 + 5;
int sz[N], m;
long long ans, ct[N], dp[N][5];
vector<int> e[N];
void dfs(int u, int fa) {
  sz[u] = dp[u][0] = 1;
  for (int v : e[u])
    if (v != fa) {
      dfs(v, u);
      ans += sz[v] * ct[u] + sz[u] * ct[v];
      for (int i = 0; i < m; i++)
        for (int j = 0; j < m; j++) ans += dp[u][i] * dp[v][j] << (i + j >= m);
      sz[u] += sz[v];
      ct[u] += ct[v] + dp[v][m - 1];
      for (int i = 0; i < m; i++) dp[u][(i + 1) % m] += dp[v][i];
    }
}
int main() {
  int n;
  rd(n);
  rd(m);
  for (int i = 1, u, v; i < n; i++)
    rd(u), rd(v), e[u].push_back(v), e[v].push_back(u);
  dfs(1, 0);
  printf("%I64d", ans);
  return 0;
}

### Prompt

Your task is to create 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 void rd(_T &_a) {
  int _f = 0, _ch = getchar();
  _a = 0;
  while (_ch < '0' || _ch > '9') {
    if (_ch == '-') _f = 1;
    _ch = getchar();
  }
  while (_ch >= '0' && _ch <= '9') _a = _a * 10 + _ch - '0', _ch = getchar();
  if (_f) _a = -_a;
}
template <class _T>
inline _T dmin(const _T _a, const _T _b) {
  return _a < _b ? _a : _b;
}
template <class _T>
inline _T dmax(const _T _a, const _T _b) {
  return _a > _b ? _a : _b;
}
template <class _T>
inline bool cmin(_T &_a, const _T _b) {
  return _a > _b ? _a = _b, 1 : 0;
}
template <class _T>
inline bool cmax(_T &_a, const _T _b) {
  return _a < _b ? _a = _b, 1 : 0;
}
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int N = 2e5 + 5;
int sz[N], m;
long long ans, ct[N], dp[N][5];
vector<int> e[N];
void dfs(int u, int fa) {
  sz[u] = dp[u][0] = 1;
  for (int v : e[u])
    if (v != fa) {
      dfs(v, u);
      ans += sz[v] * ct[u] + sz[u] * ct[v];
      for (int i = 0; i < m; i++)
        for (int j = 0; j < m; j++) ans += dp[u][i] * dp[v][j] << (i + j >= m);
      sz[u] += sz[v];
      ct[u] += ct[v] + dp[v][m - 1];
      for (int i = 0; i < m; i++) dp[u][(i + 1) % m] += dp[v][i];
    }
}
int main() {
  int n;
  rd(n);
  rd(m);
  for (int i = 1, u, v; i < n; i++)
    rd(u), rd(v), e[u].push_back(v), e[v].push_back(u);
  dfs(1, 0);
  printf("%I64d", ans);
  return 0;
}

```