#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const int nmax = 200200;
const int kmax = 6;
int n, k;
vector<int> g[nmax];
long long cnt[nmax][kmax], sumLen[nmax][kmax];
long long res;
int used[nmax];
void solve(int v, int p) {
  used[v] = 1;
  sumLen[v][0] = 0;
  cnt[v][0] = 1;
  for (int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) {
      int u = g[v][i];
      solve(u, v);
      for (int j = 0; j < k; j++)
        for (int e = 0; e < k; e++) {
          res += sumLen[u][j] * cnt[v][e];
          res += sumLen[v][e] * cnt[u][j];
          if (e + j + 1 > 0) res += cnt[u][j] * cnt[v][e];
          if (e + j + 1 > k) res += cnt[u][j] * cnt[v][e];
        }
      for (int j = 0; j < k - 1; j++) {
        sumLen[v][j + 1] += sumLen[u][j];
        cnt[v][j + 1] += cnt[u][j];
      }
      sumLen[v][0] += sumLen[u][k - 1] + cnt[u][k - 1];
      cnt[v][0] += cnt[u][k - 1];
    }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  solve(0, -1);
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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;
const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const int nmax = 200200;
const int kmax = 6;
int n, k;
vector<int> g[nmax];
long long cnt[nmax][kmax], sumLen[nmax][kmax];
long long res;
int used[nmax];
void solve(int v, int p) {
  used[v] = 1;
  sumLen[v][0] = 0;
  cnt[v][0] = 1;
  for (int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) {
      int u = g[v][i];
      solve(u, v);
      for (int j = 0; j < k; j++)
        for (int e = 0; e < k; e++) {
          res += sumLen[u][j] * cnt[v][e];
          res += sumLen[v][e] * cnt[u][j];
          if (e + j + 1 > 0) res += cnt[u][j] * cnt[v][e];
          if (e + j + 1 > k) res += cnt[u][j] * cnt[v][e];
        }
      for (int j = 0; j < k - 1; j++) {
        sumLen[v][j + 1] += sumLen[u][j];
        cnt[v][j + 1] += cnt[u][j];
      }
      sumLen[v][0] += sumLen[u][k - 1] + cnt[u][k - 1];
      cnt[v][0] += cnt[u][k - 1];
    }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  solve(0, -1);
  cout << res << endl;
  return 0;
}

```