#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const double pi = acos(1.0 * (-1));
struct node {
  int p;
  vector<int> v;
} ns[200022];
long long sn[200022][5];
long long ss[200022];
long long sum = 0;
int n, k;
void init(int now, int p) {
  ss[now]++;
  sn[now][0]++;
  for (int i = 0; i < ns[now].v.size(); i++) {
    if (ns[now].v[i] == p) continue;
    init(ns[now].v[i], now);
    ss[now] += ss[ns[now].v[i]];
    for (int i1 = 0; i1 < k; i1++) {
      for (int i2 = 0; i2 < k; i2++) {
        int f = (i1 + i2 - 1 + k) % k;
        sum += sn[now][i1] * sn[ns[now].v[i]][i2] * f;
      }
    }
    for (int j = 0; j < k; j++) sn[now][(k + j - 1) % k] += sn[ns[now].v[i]][j];
    sum += ss[ns[now].v[i]] * (n - ss[ns[now].v[i]]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int l, r;
    cin >> l >> r;
    ns[l].v.push_back(r);
    ns[r].v.push_back(l);
  }
  init(1, -1);
  cout << sum / k;
  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 long long mod = 1000000007;
const double pi = acos(1.0 * (-1));
struct node {
  int p;
  vector<int> v;
} ns[200022];
long long sn[200022][5];
long long ss[200022];
long long sum = 0;
int n, k;
void init(int now, int p) {
  ss[now]++;
  sn[now][0]++;
  for (int i = 0; i < ns[now].v.size(); i++) {
    if (ns[now].v[i] == p) continue;
    init(ns[now].v[i], now);
    ss[now] += ss[ns[now].v[i]];
    for (int i1 = 0; i1 < k; i1++) {
      for (int i2 = 0; i2 < k; i2++) {
        int f = (i1 + i2 - 1 + k) % k;
        sum += sn[now][i1] * sn[ns[now].v[i]][i2] * f;
      }
    }
    for (int j = 0; j < k; j++) sn[now][(k + j - 1) % k] += sn[ns[now].v[i]][j];
    sum += ss[ns[now].v[i]] * (n - ss[ns[now].v[i]]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int l, r;
    cin >> l >> r;
    ns[l].v.push_back(r);
    ns[r].v.push_back(l);
  }
  init(1, -1);
  cout << sum / k;
  return 0;
}

```