#include <bits/stdc++.h>
using namespace std;
const int NX = 200010;
vector<int> adj[NX];
long long S[NX], Ni[NX][6], Si[NX][6];
int N, K;
void print(int x) {
  cout << "Node " << x << endl;
  cout << " " << S[x] << endl;
  for (int i = 0; i < K; i++) cout << " " << Ni[x][i];
  cout << endl;
  for (int i = 0; i < K; i++) cout << " " << Si[x][i];
  cout << "\n\n";
}
void merge(int x, int y) {
  S[x] += S[y];
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++) {
      long long st = S[x], t = (i + j + 1) % K, ss = 0;
      ss += Si[x][i] * Ni[y][j];
      ss += (Si[y][j] + Ni[y][j]) * Ni[x][i];
      ss -= t * Ni[x][i] * Ni[y][j];
      S[x] += ss / K;
      if (t) S[x] += Ni[x][i] * Ni[y][j];
    }
  for (int i = 0; i < K; i++) {
    Ni[x][(i + 1) % K] += Ni[y][i];
    Si[x][(i + 1) % K] += Si[y][i] + Ni[y][i];
  }
}
long long sol(int p, int x) {
  Ni[x][0] = 1;
  for (auto y : adj[x])
    if (y != p) {
      sol(x, y);
      merge(x, y);
    }
  return S[x];
}
int main() {
  cin >> N >> K;
  for (int a, b, i = 1; i < N; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  cout << sol(1, 1) << endl;
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int NX = 200010;
vector<int> adj[NX];
long long S[NX], Ni[NX][6], Si[NX][6];
int N, K;
void print(int x) {
  cout << "Node " << x << endl;
  cout << " " << S[x] << endl;
  for (int i = 0; i < K; i++) cout << " " << Ni[x][i];
  cout << endl;
  for (int i = 0; i < K; i++) cout << " " << Si[x][i];
  cout << "\n\n";
}
void merge(int x, int y) {
  S[x] += S[y];
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++) {
      long long st = S[x], t = (i + j + 1) % K, ss = 0;
      ss += Si[x][i] * Ni[y][j];
      ss += (Si[y][j] + Ni[y][j]) * Ni[x][i];
      ss -= t * Ni[x][i] * Ni[y][j];
      S[x] += ss / K;
      if (t) S[x] += Ni[x][i] * Ni[y][j];
    }
  for (int i = 0; i < K; i++) {
    Ni[x][(i + 1) % K] += Ni[y][i];
    Si[x][(i + 1) % K] += Si[y][i] + Ni[y][i];
  }
}
long long sol(int p, int x) {
  Ni[x][0] = 1;
  for (auto y : adj[x])
    if (y != p) {
      sol(x, y);
      merge(x, y);
    }
  return S[x];
}
int main() {
  cin >> N >> K;
  for (int a, b, i = 1; i < N; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  cout << sol(1, 1) << endl;
  return 0;
}

```