#include <bits/stdc++.h>
using namespace std;
const long long N = 2 * 1e5 + 9;
const long long oo = 1e15;
const long long mod = 1e9 + 7;
long long n, k, sum;
long long dp[N][6], child[N];
vector<long long> vi[N];
void dfs(long long pre, long long u) {
  child[u] = 1;
  dp[u][0] = 1;
  for (long long v : vi[u]) {
    if (v != pre) {
      dfs(u, v);
      child[u] += child[v];
      sum += child[v] * (n - child[v]);
      for (long long j = 0; j <= k - 1; ++j)
        for (long long t = 0; t <= k - 1; ++t)
          if ((j + 1 + t) % k != 0)
            sum += dp[u][j] * dp[v][t] * (k - (j + t + 1) % k);
      for (long long j = 0; j < k; ++j) dp[u][(j + 1) % k] += dp[v][j];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    long long x, y;
    cin >> x >> y;
    vi[x].push_back(y);
    vi[y].push_back(x);
  }
  dfs(1, 1);
  cout << sum / k;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 N = 2 * 1e5 + 9;
const long long oo = 1e15;
const long long mod = 1e9 + 7;
long long n, k, sum;
long long dp[N][6], child[N];
vector<long long> vi[N];
void dfs(long long pre, long long u) {
  child[u] = 1;
  dp[u][0] = 1;
  for (long long v : vi[u]) {
    if (v != pre) {
      dfs(u, v);
      child[u] += child[v];
      sum += child[v] * (n - child[v]);
      for (long long j = 0; j <= k - 1; ++j)
        for (long long t = 0; t <= k - 1; ++t)
          if ((j + 1 + t) % k != 0)
            sum += dp[u][j] * dp[v][t] * (k - (j + t + 1) % k);
      for (long long j = 0; j < k; ++j) dp[u][(j + 1) % k] += dp[v][j];
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    long long x, y;
    cin >> x >> y;
    vi[x].push_back(y);
    vi[y].push_back(x);
  }
  dfs(1, 1);
  cout << sum / k;
}

```