#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void nope(int dec = 0) {
  if (!dec)
    cout << "NO";
  else
    cout << "YES";
  exit(0);
}
long long n, k;
vector<int> adj_list[2000100];
long long mods[2000100][5];
long long sum[2000100];
bool mark[20001000];
long long S = 0, F = 0;
void solve(int node) {
  if (mark[node]) return;
  mark[node] = 1;
  long long curr[6];
  memset(curr, 0, sizeof(curr));
  for (auto child : adj_list[node]) {
    if (mark[child]) continue;
    solve(child);
    S += sum[child] * (n - sum[child]);
    sum[node] += sum[child];
    for (int j = 0; j < k; j++) curr[j] += mods[child][j];
    mark[child] = 0;
  }
  sum[node]++;
  for (int i = 0; i < k; i++) mods[node][(i + 1) % k] = curr[i];
  for (int i = 0; i < k; i++) F += ((k - i) % k) * mods[node][i];
  for (auto child : adj_list[node]) {
    if (mark[child]) continue;
    for (int i = 0; i < k; i++) curr[i] -= mods[child][i];
    for (int i = 0; i < k; i++)
      for (int j = 0; j < k; j++) {
        long long cnt = 0;
        int val = (i + j + 2) % k;
        cnt = mods[child][i] * (curr[j]);
        F += ((k - val) % k) * cnt;
      }
    mark[child] = 1;
  }
  mods[node][0]++;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj_list[a].push_back(b);
    adj_list[b].push_back(a);
  }
  solve(1);
  cout << (S + F) / k << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void nope(int dec = 0) {
  if (!dec)
    cout << "NO";
  else
    cout << "YES";
  exit(0);
}
long long n, k;
vector<int> adj_list[2000100];
long long mods[2000100][5];
long long sum[2000100];
bool mark[20001000];
long long S = 0, F = 0;
void solve(int node) {
  if (mark[node]) return;
  mark[node] = 1;
  long long curr[6];
  memset(curr, 0, sizeof(curr));
  for (auto child : adj_list[node]) {
    if (mark[child]) continue;
    solve(child);
    S += sum[child] * (n - sum[child]);
    sum[node] += sum[child];
    for (int j = 0; j < k; j++) curr[j] += mods[child][j];
    mark[child] = 0;
  }
  sum[node]++;
  for (int i = 0; i < k; i++) mods[node][(i + 1) % k] = curr[i];
  for (int i = 0; i < k; i++) F += ((k - i) % k) * mods[node][i];
  for (auto child : adj_list[node]) {
    if (mark[child]) continue;
    for (int i = 0; i < k; i++) curr[i] -= mods[child][i];
    for (int i = 0; i < k; i++)
      for (int j = 0; j < k; j++) {
        long long cnt = 0;
        int val = (i + j + 2) % k;
        cnt = mods[child][i] * (curr[j]);
        F += ((k - val) % k) * cnt;
      }
    mark[child] = 1;
  }
  mods[node][0]++;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj_list[a].push_back(b);
    adj_list[b].push_back(a);
  }
  solve(1);
  cout << (S + F) / k << endl;
  return 0;
}

```