#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
int n, k;
vector<int> edges[200005];
bool v[200005];
int c[200005];
int p[200005][5];
long long int res;
void dfs(int cur, int depth) {
  v[cur] = 1;
  int nxt, tmp;
  p[cur][depth % k] = 1;
  c[cur] = 1;
  for (int i = 0; i < edges[cur].size(); ++i) {
    nxt = edges[cur][i];
    if (!v[nxt]) {
      dfs(nxt, depth + 1);
      for (int a = 0; a < k; ++a) {
        for (int b = 0; b < k; ++b) {
          tmp = ((k << 1) + a + b - ((depth % k) << 1)) % k;
          tmp = ((k << 1) - tmp) % k;
          res += (long long int)tmp * p[cur][a] * p[nxt][b];
        }
      }
      for (int j = 0; j < k; ++j) p[cur][j] += p[nxt][j];
      c[cur] += c[nxt];
    }
  }
  res += (long long int)c[cur] * (n - c[cur]);
}
int main() {
  ios_base::sync_with_stdio(0);
  int a, b;
  while (cin >> n >> k) {
    for (int i = 0; i < n - 1; ++i) {
      cin >> a >> b;
      edges[a].push_back(b);
      edges[b].push_back(a);
    }
    res = 0;
    memset(v, 0, sizeof(v));
    memset(p, 0, sizeof(p));
    dfs(1, 1);
    cout << res / k << "\n";
    for (int i = 1; i <= n; ++i) edges[i].clear();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 inf = 0x3f3f3f3f;
const double eps = 1e-8;
int n, k;
vector<int> edges[200005];
bool v[200005];
int c[200005];
int p[200005][5];
long long int res;
void dfs(int cur, int depth) {
  v[cur] = 1;
  int nxt, tmp;
  p[cur][depth % k] = 1;
  c[cur] = 1;
  for (int i = 0; i < edges[cur].size(); ++i) {
    nxt = edges[cur][i];
    if (!v[nxt]) {
      dfs(nxt, depth + 1);
      for (int a = 0; a < k; ++a) {
        for (int b = 0; b < k; ++b) {
          tmp = ((k << 1) + a + b - ((depth % k) << 1)) % k;
          tmp = ((k << 1) - tmp) % k;
          res += (long long int)tmp * p[cur][a] * p[nxt][b];
        }
      }
      for (int j = 0; j < k; ++j) p[cur][j] += p[nxt][j];
      c[cur] += c[nxt];
    }
  }
  res += (long long int)c[cur] * (n - c[cur]);
}
int main() {
  ios_base::sync_with_stdio(0);
  int a, b;
  while (cin >> n >> k) {
    for (int i = 0; i < n - 1; ++i) {
      cin >> a >> b;
      edges[a].push_back(b);
      edges[b].push_back(a);
    }
    res = 0;
    memset(v, 0, sizeof(v));
    memset(p, 0, sizeof(p));
    dfs(1, 1);
    cout << res / k << "\n";
    for (int i = 1; i <= n; ++i) edges[i].clear();
  }
  return 0;
}

```