#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 10;
const int MAXN = 1e4 + 10;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
vector<int> V[200100];
unsigned long long F[200100], G[200100][10], cnt[200100], sum[200100];
int n, k;
void DFS(int x, int father) {
  cnt[x] = 1LL;
  G[x][0] = 1LL;
  for (int i = 0; i < V[x].size(); i++) {
    int t = V[x][i];
    if (t == father) continue;
    DFS(t, x);
    F[x] += F[t];
    F[x] += sum[x] * cnt[t] + sum[t] * cnt[x];
    for (int p = 0; p < k; p++)
      for (int q = 0; q < k; q++) {
        unsigned long long tmp = G[x][p] * G[t][q];
        if (p + q > k - 1) tmp *= 2LL;
        F[x] += tmp;
      }
    for (int j = 0; j < k; j++) G[x][(j + 1) % k] += G[t][j];
    sum[x] += sum[t];
    cnt[x] += cnt[t];
    sum[x] += G[t][k - 1];
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    V[u].push_back(v);
    V[v].push_back(u);
  }
  unsigned long long res = 0;
  DFS(1, -1);
  cout << F[1];
  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;
const int MAX = 1e5 + 10;
const int MAXN = 1e4 + 10;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
vector<int> V[200100];
unsigned long long F[200100], G[200100][10], cnt[200100], sum[200100];
int n, k;
void DFS(int x, int father) {
  cnt[x] = 1LL;
  G[x][0] = 1LL;
  for (int i = 0; i < V[x].size(); i++) {
    int t = V[x][i];
    if (t == father) continue;
    DFS(t, x);
    F[x] += F[t];
    F[x] += sum[x] * cnt[t] + sum[t] * cnt[x];
    for (int p = 0; p < k; p++)
      for (int q = 0; q < k; q++) {
        unsigned long long tmp = G[x][p] * G[t][q];
        if (p + q > k - 1) tmp *= 2LL;
        F[x] += tmp;
      }
    for (int j = 0; j < k; j++) G[x][(j + 1) % k] += G[t][j];
    sum[x] += sum[t];
    cnt[x] += cnt[t];
    sum[x] += G[t][k - 1];
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    V[u].push_back(v);
    V[v].push_back(u);
  }
  unsigned long long res = 0;
  DFS(1, -1);
  cout << F[1];
  return 0;
}

```