#include <bits/stdc++.h>
using namespace std;
template <typename G>
inline void read(G &x) {
  x = 0;
  char c;
  while (!isdigit(c = getchar()))
    ;
  do {
    x = x * 10 + c - '0';
  } while (isdigit(c = getchar()));
}
template <typename G>
inline void write(G x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(a, min(b, c, d));
}
template <class T>
inline T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(a, max(b, c, d));
}
const int MAXN = 1e5 * 3;
const int base = 1e9 + 7;
const int N = 5000;
int n, k;
int f[MAXN][7], dp[MAXN][7];
long long ans;
long long cnt[MAXN], dem[MAXN];
vector<int> a[MAXN];
void dfs(int u, int p) {
  ++dp[u][0];
  ++f[u][0];
  long long ddem = 0;
  for (int i = 0; int v = a[u][i]; ++i)
    if (v != p) {
      cnt[v] = cnt[u] + dem[u];
      for (int j = (0); j <= (k - 1); ++j) {
        f[v][(j + 1) % k] += f[u][j];
        if ((j + 1) % k == 0)
          cnt[v] += f[u][j];
        else
          ans += f[u][j];
      }
      ans += cnt[v];
      dfs(v, u);
      ddem = 0;
      for (int j = (0); j <= (k - 1); ++j) {
        f[u][(j + 1) % k] += dp[v][j];
        dp[u][(j + 1) % k] += dp[v][j];
        if ((j + 1) % k == 0) dem[u] += dp[v][j];
      }
      dem[u] += dem[v];
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1); i < (n); ++i) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = (1); i <= (n); ++i) a[i].push_back(0);
  dfs(1, -1);
  cout << ans << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 G>
inline void read(G &x) {
  x = 0;
  char c;
  while (!isdigit(c = getchar()))
    ;
  do {
    x = x * 10 + c - '0';
  } while (isdigit(c = getchar()));
}
template <typename G>
inline void write(G x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(a, min(b, c, d));
}
template <class T>
inline T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(a, max(b, c, d));
}
const int MAXN = 1e5 * 3;
const int base = 1e9 + 7;
const int N = 5000;
int n, k;
int f[MAXN][7], dp[MAXN][7];
long long ans;
long long cnt[MAXN], dem[MAXN];
vector<int> a[MAXN];
void dfs(int u, int p) {
  ++dp[u][0];
  ++f[u][0];
  long long ddem = 0;
  for (int i = 0; int v = a[u][i]; ++i)
    if (v != p) {
      cnt[v] = cnt[u] + dem[u];
      for (int j = (0); j <= (k - 1); ++j) {
        f[v][(j + 1) % k] += f[u][j];
        if ((j + 1) % k == 0)
          cnt[v] += f[u][j];
        else
          ans += f[u][j];
      }
      ans += cnt[v];
      dfs(v, u);
      ddem = 0;
      for (int j = (0); j <= (k - 1); ++j) {
        f[u][(j + 1) % k] += dp[v][j];
        dp[u][(j + 1) % k] += dp[v][j];
        if ((j + 1) % k == 0) dem[u] += dp[v][j];
      }
      dem[u] += dem[v];
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1); i < (n); ++i) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = (1); i <= (n); ++i) a[i].push_back(0);
  dfs(1, -1);
  cout << ans << '\n';
  return 0;
}

```