#include <bits/stdc++.h>
const int INF = 1e9;
const long long lINF = 9223372036854775807;
using namespace std;
ifstream in;
ofstream out;
vector<bool> used;
vector<vector<long long>> g;
vector<long long> pl(5);
long long n, k, answ;
vector<long long> dfs(long long v, long long lv) {
  vector<long long> v0(k + 1, 0);
  v0[k] = v0[0] = 1;
  for (auto e : g[v])
    if (e != lv) {
      vector<long long> v1 = dfs(e, v);
      vector<long long> v2(k);
      for (int i = 0; i < (k); i++) {
        for (int j = (0); j < (k); j++) pl[(i + 1 + j) % k] += v1[i] * v0[j];
      }
      for (int i = 0; i < (k); i++) v0[(i + 1) % k] += v1[i];
      v0[k] += v1[k];
      answ += v1[k] * (n - v1[k]);
    }
  return v0;
}
int main() {
  in.open("input.txt");
  out.open("output.txt");
  cin >> n >> k;
  g.resize(n);
  for (int i = 0; i < (n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  for (int i = 0; i < (5); i++) answ += pl[i] * ((k - i) % k);
  cout << answ / k;
  return 0;
}

### Prompt

Your challenge is to write 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>
const int INF = 1e9;
const long long lINF = 9223372036854775807;
using namespace std;
ifstream in;
ofstream out;
vector<bool> used;
vector<vector<long long>> g;
vector<long long> pl(5);
long long n, k, answ;
vector<long long> dfs(long long v, long long lv) {
  vector<long long> v0(k + 1, 0);
  v0[k] = v0[0] = 1;
  for (auto e : g[v])
    if (e != lv) {
      vector<long long> v1 = dfs(e, v);
      vector<long long> v2(k);
      for (int i = 0; i < (k); i++) {
        for (int j = (0); j < (k); j++) pl[(i + 1 + j) % k] += v1[i] * v0[j];
      }
      for (int i = 0; i < (k); i++) v0[(i + 1) % k] += v1[i];
      v0[k] += v1[k];
      answ += v1[k] * (n - v1[k]);
    }
  return v0;
}
int main() {
  in.open("input.txt");
  out.open("output.txt");
  cin >> n >> k;
  g.resize(n);
  for (int i = 0; i < (n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  for (int i = 0; i < (5); i++) answ += pl[i] * ((k - i) % k);
  cout << answ / k;
  return 0;
}

```