#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > way;
int n, k;
long long cnt[1000000], dp[1000000][6];
long long ans;
int fun(int aa, int bb) { return ((aa - bb) % k + k) % k; }
void dfs(int now, int prd, int dist) {
  cnt[now] = dp[now][dist % k] = 1;
  for (auto to : way[now])
    if (to != prd) {
      dfs(to, now, dist + 1);
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
          ans += fun(k, fun(i + j, 2 * dist)) * dp[now][i] * dp[to][j];
      for (int i = 0; i < k; i++) dp[now][i] += dp[to][i];
      cnt[now] += cnt[to];
    }
  ans += cnt[now] * (n - cnt[now]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  way.resize(n + 1);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    way[a].push_back(b);
    way[b].push_back(a);
  }
  dfs(1, -1, 0);
  cout << ans / k;
}

### Prompt

Generate 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;
vector<vector<int> > way;
int n, k;
long long cnt[1000000], dp[1000000][6];
long long ans;
int fun(int aa, int bb) { return ((aa - bb) % k + k) % k; }
void dfs(int now, int prd, int dist) {
  cnt[now] = dp[now][dist % k] = 1;
  for (auto to : way[now])
    if (to != prd) {
      dfs(to, now, dist + 1);
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
          ans += fun(k, fun(i + j, 2 * dist)) * dp[now][i] * dp[to][j];
      for (int i = 0; i < k; i++) dp[now][i] += dp[to][i];
      cnt[now] += cnt[to];
    }
  ans += cnt[now] * (n - cnt[now]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  way.resize(n + 1);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    way[a].push_back(b);
    way[b].push_back(a);
  }
  dfs(1, -1, 0);
  cout << ans / k;
}

```