#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
const int K = 5;
int n, k;
int dp[N][K];
vector<int> G[N];
long long dfs(int u, int p, int d) {
  dp[u][d % k] = 1;
  long long ans = 0;
  int dk = k - (d % k);
  for (int &v : G[u]) {
    if (v == p) continue;
    ans += dfs(v, u, d + 1);
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        ans += 1LL * ((k - ((i + j + dk + dk) % k)) % k) * dp[u][i] * dp[v][j];
    for (int i = 0; i < k; ++i) dp[u][i] += dp[v][i];
  }
  int t = 0;
  for (int i = 0; i < k; ++i) t += dp[u][i];
  ans += 1LL * t * (n - t);
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  while (cin >> n >> k) {
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 1, u, v; i < n; ++i) {
      cin >> u >> v;
      G[u].push_back(v);
      G[v].push_back(u);
    }
    memset(dp, 0, sizeof(dp));
    cout << dfs(1, 1, 0) / k << ('\n');
  }
}

### Prompt

Please formulate 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;
const int N = 1 << 18;
const int K = 5;
int n, k;
int dp[N][K];
vector<int> G[N];
long long dfs(int u, int p, int d) {
  dp[u][d % k] = 1;
  long long ans = 0;
  int dk = k - (d % k);
  for (int &v : G[u]) {
    if (v == p) continue;
    ans += dfs(v, u, d + 1);
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j)
        ans += 1LL * ((k - ((i + j + dk + dk) % k)) % k) * dp[u][i] * dp[v][j];
    for (int i = 0; i < k; ++i) dp[u][i] += dp[v][i];
  }
  int t = 0;
  for (int i = 0; i < k; ++i) t += dp[u][i];
  ans += 1LL * t * (n - t);
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  while (cin >> n >> k) {
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 1, u, v; i < n; ++i) {
      cin >> u >> v;
      G[u].push_back(v);
      G[v].push_back(u);
    }
    memset(dp, 0, sizeof(dp));
    cout << dfs(1, 1, 0) / k << ('\n');
  }
}

```