#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& a) {
  out << "[" << a.first << " " << a.second << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const set<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const multiset<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U>& a) {
  for (auto& it : a) out << it.first << " -> " << it.second << " | ";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, const multimap<T, U>& a) {
  for (auto& it : a) out << it.first << " -> " << it.second << " | ";
  return out;
}
int x_dir[4] = {-1, 0, 1, 0};
int y_dir[4] = {0, 1, 0, -1};
const int max_n = 2e5 + 10;
int n, k;
vector<vector<int> > adj;
long long ans;
long long dp[max_n][6];
long long subtrees[max_n];
long long sumpaths[6];
void dfs(int u, int parent) {
  subtrees[u] = 1;
  dp[u][0] = 1;
  for (auto v : adj[u]) {
    if (v == parent) continue;
    dfs(v, u);
    for (int i = int(0); i <= int(k - 1); i++)
      for (int j = int(0); j <= int(k - 1); j++)
        sumpaths[(i + j + 1) % k] += dp[u][i] * dp[v][j];
    for (int i = int(0); i <= int(k - 1); i++)
      dp[u][i] += dp[v][(i + k - 1) % k];
    ans += subtrees[v] * (n - subtrees[v]);
    subtrees[u] += subtrees[v];
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  adj.assign(n + 1, vector<int>());
  for (int i = int(1); i <= int(n - 1); i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  for (int i = int(1); i <= int(k - 1); i++) ans += (k - i) * sumpaths[i];
  cout << ans / k;
  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 <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& a) {
  out << "[" << a.first << " " << a.second << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const set<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const multiset<T>& a) {
  out << "[ ";
  for (auto& it : a) out << it << " ";
  out << "]";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U>& a) {
  for (auto& it : a) out << it.first << " -> " << it.second << " | ";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, const multimap<T, U>& a) {
  for (auto& it : a) out << it.first << " -> " << it.second << " | ";
  return out;
}
int x_dir[4] = {-1, 0, 1, 0};
int y_dir[4] = {0, 1, 0, -1};
const int max_n = 2e5 + 10;
int n, k;
vector<vector<int> > adj;
long long ans;
long long dp[max_n][6];
long long subtrees[max_n];
long long sumpaths[6];
void dfs(int u, int parent) {
  subtrees[u] = 1;
  dp[u][0] = 1;
  for (auto v : adj[u]) {
    if (v == parent) continue;
    dfs(v, u);
    for (int i = int(0); i <= int(k - 1); i++)
      for (int j = int(0); j <= int(k - 1); j++)
        sumpaths[(i + j + 1) % k] += dp[u][i] * dp[v][j];
    for (int i = int(0); i <= int(k - 1); i++)
      dp[u][i] += dp[v][(i + k - 1) % k];
    ans += subtrees[v] * (n - subtrees[v]);
    subtrees[u] += subtrees[v];
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  adj.assign(n + 1, vector<int>());
  for (int i = int(1); i <= int(n - 1); i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  for (int i = int(1); i <= int(k - 1); i++) ans += (k - i) * sumpaths[i];
  cout << ans / k;
  return 0;
}

```