#include <bits/stdc++.h>
using namespace std;
template <typename T>
void rd(T& x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) f ^= !(c ^ 45);
  x = (c & 15);
  while (isdigit(c = getchar())) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
void pt(T x, int c = 10) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10, -1);
  putchar(x % 10 + 48);
  if (c != -1) putchar(c);
}
template <typename T>
void umax(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
void umin(T& x, const T& y) {
  if (x > y) x = y;
}
using vi = vector<int>;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long, long>;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int n, k;
vector<int> g[maxn];
ll f[maxn];
int siz[maxn];
int h[maxn][5];
ll sum = 0;
void dfs(int u, int pre) {
  siz[u] = 1;
  h[u][0] = 1;
  f[u] = 0;
  for (auto v : g[u]) {
    if (v == pre) continue;
    dfs(v, u);
    int t[5];
    ll ff = f[v] + h[v][k - 1];
    for (int i = 1; i < k; i++) t[i] = h[v][i - 1];
    t[0] = h[v][k - 1];
    sum += f[u] * siz[v] + ff * siz[u];
    for (int i = 0; i < k; i++)
      for (int j = 0; j < k; j++) {
        if (i + j == 0) continue;
        if (i + j <= k)
          sum += 1LL * h[u][i] * t[j];
        else
          sum += 2LL * h[u][i] * t[j];
      }
    siz[u] += siz[v];
    f[u] += ff;
    for (int i = 0; i < k; i++) h[u][i] += t[i];
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cout << sum << "\n";
  return 0;
}

### Prompt

Create a solution in cpp for 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 <typename T>
void rd(T& x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) f ^= !(c ^ 45);
  x = (c & 15);
  while (isdigit(c = getchar())) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
void pt(T x, int c = 10) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10, -1);
  putchar(x % 10 + 48);
  if (c != -1) putchar(c);
}
template <typename T>
void umax(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
void umin(T& x, const T& y) {
  if (x > y) x = y;
}
using vi = vector<int>;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long, long>;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int n, k;
vector<int> g[maxn];
ll f[maxn];
int siz[maxn];
int h[maxn][5];
ll sum = 0;
void dfs(int u, int pre) {
  siz[u] = 1;
  h[u][0] = 1;
  f[u] = 0;
  for (auto v : g[u]) {
    if (v == pre) continue;
    dfs(v, u);
    int t[5];
    ll ff = f[v] + h[v][k - 1];
    for (int i = 1; i < k; i++) t[i] = h[v][i - 1];
    t[0] = h[v][k - 1];
    sum += f[u] * siz[v] + ff * siz[u];
    for (int i = 0; i < k; i++)
      for (int j = 0; j < k; j++) {
        if (i + j == 0) continue;
        if (i + j <= k)
          sum += 1LL * h[u][i] * t[j];
        else
          sum += 2LL * h[u][i] * t[j];
      }
    siz[u] += siz[v];
    f[u] += ff;
    for (int i = 0; i < k; i++) h[u][i] += t[i];
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cout << sum << "\n";
  return 0;
}

```