#include <bits/stdc++.h>
using namespace std;
template <class BidirIt>
BidirIt prev(BidirIt it,
             typename iterator_traits<BidirIt>::difference_type n = 1) {
  advance(it, -n);
  return it;
}
template <class ForwardIt>
ForwardIt next(ForwardIt it,
               typename iterator_traits<ForwardIt>::difference_type n = 1) {
  advance(it, n);
  return it;
}
const double EPS = 1e-9;
const double PI = 3.141592653589793238462;
template <typename T>
inline T sq(T a) {
  return a * a;
}
const int MAX_N = 2e5 + 5;
const int MAX_K = 6;
int n, k;
long long down[MAX_N][MAX_K], up[MAX_N][MAX_K];
long long dist_d[MAX_N], dist_u[MAX_N];
long long sz[MAX_N];
vector<int> gr[MAX_N];
void init(int u, int par) {
  sz[u] = 1;
  for (auto v : gr[u])
    if (v != par) {
      init(v, u);
      sz[u] += sz[v];
      dist_d[u] += dist_d[v] + sz[v];
    }
  down[u][0]++;
  for (int i = 0; i < k; i++)
    for (auto v : gr[u]) down[u][i] += down[v][((i) + k + (-1)) % k];
}
void dfs(int u, int par) {
  if (par != -1)
    dist_u[u] = dist_u[par] + dist_d[par] - (dist_d[u] + sz[u]) + n - sz[u];
  up[u][0]++;
  if (par != -1) {
    for (int i = 0; i < k; i++) {
      up[u][i] +=
          up[par][((i) + k + (-1)) % k] +
          (down[par][((i) + k + (-1)) % k] - down[u][((i) + k + (-2)) % k]);
    }
    up[u][1 % k]--;
  }
  for (auto v : gr[u])
    if (v != par) dfs(v, u);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  init(1, -1);
  dfs(1, -1);
  long long ans = 0;
  for (int u = 1; u <= n; u++) {
    ans += dist_u[u] + dist_d[u];
    for (int i = 1; i < k; i++) ans += (down[u][i] + up[u][i]) * 1LL * (k - i);
  }
  ans /= 2LL * k;
  printf("%lld\n", ans);
  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 <class BidirIt>
BidirIt prev(BidirIt it,
             typename iterator_traits<BidirIt>::difference_type n = 1) {
  advance(it, -n);
  return it;
}
template <class ForwardIt>
ForwardIt next(ForwardIt it,
               typename iterator_traits<ForwardIt>::difference_type n = 1) {
  advance(it, n);
  return it;
}
const double EPS = 1e-9;
const double PI = 3.141592653589793238462;
template <typename T>
inline T sq(T a) {
  return a * a;
}
const int MAX_N = 2e5 + 5;
const int MAX_K = 6;
int n, k;
long long down[MAX_N][MAX_K], up[MAX_N][MAX_K];
long long dist_d[MAX_N], dist_u[MAX_N];
long long sz[MAX_N];
vector<int> gr[MAX_N];
void init(int u, int par) {
  sz[u] = 1;
  for (auto v : gr[u])
    if (v != par) {
      init(v, u);
      sz[u] += sz[v];
      dist_d[u] += dist_d[v] + sz[v];
    }
  down[u][0]++;
  for (int i = 0; i < k; i++)
    for (auto v : gr[u]) down[u][i] += down[v][((i) + k + (-1)) % k];
}
void dfs(int u, int par) {
  if (par != -1)
    dist_u[u] = dist_u[par] + dist_d[par] - (dist_d[u] + sz[u]) + n - sz[u];
  up[u][0]++;
  if (par != -1) {
    for (int i = 0; i < k; i++) {
      up[u][i] +=
          up[par][((i) + k + (-1)) % k] +
          (down[par][((i) + k + (-1)) % k] - down[u][((i) + k + (-2)) % k]);
    }
    up[u][1 % k]--;
  }
  for (auto v : gr[u])
    if (v != par) dfs(v, u);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  init(1, -1);
  dfs(1, -1);
  long long ans = 0;
  for (int u = 1; u <= n; u++) {
    ans += dist_u[u] + dist_d[u];
    for (int i = 1; i < k; i++) ans += (down[u][i] + up[u][i]) * 1LL * (k - i);
  }
  ans /= 2LL * k;
  printf("%lld\n", ans);
  return 0;
}

```