#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 5e5 + 10, M = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
vector<int> G[N];
long long dp[N][10], fs[N][10];
long long ans = 0;
int n, k, a, b;
void dfs(int u, int f) {
  dp[u][0] = 1;
  for (int i = 0; i < G[u].size(); ++i) {
    int to = G[u][i];
    if (to == f) continue;
    dfs(to, u);
    for (int j = 0; j <= k; ++j) {
      for (int h = 0; h <= k; ++h) {
        if (!dp[u][j] || !dp[to][h]) continue;
        long long tmp = j + h + 1;
        if (tmp % k)
          tmp = tmp / k + 1;
        else
          tmp = tmp / k;
        ans += tmp * dp[to][h] * dp[u][j] + fs[u][j] * dp[to][h] +
               fs[to][h] * dp[u][j];
      }
    }
    for (int j = 1; j <= k; ++j)
      dp[u][j] += dp[to][j - 1], fs[u][j] += fs[to][j - 1];
    dp[u][1] += dp[to][k];
    fs[u][1] += fs[to][k] + dp[to][k];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 5e5 + 10, M = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
vector<int> G[N];
long long dp[N][10], fs[N][10];
long long ans = 0;
int n, k, a, b;
void dfs(int u, int f) {
  dp[u][0] = 1;
  for (int i = 0; i < G[u].size(); ++i) {
    int to = G[u][i];
    if (to == f) continue;
    dfs(to, u);
    for (int j = 0; j <= k; ++j) {
      for (int h = 0; h <= k; ++h) {
        if (!dp[u][j] || !dp[to][h]) continue;
        long long tmp = j + h + 1;
        if (tmp % k)
          tmp = tmp / k + 1;
        else
          tmp = tmp / k;
        ans += tmp * dp[to][h] * dp[u][j] + fs[u][j] * dp[to][h] +
               fs[to][h] * dp[u][j];
      }
    }
    for (int j = 1; j <= k; ++j)
      dp[u][j] += dp[to][j - 1], fs[u][j] += fs[to][j - 1];
    dp[u][1] += dp[to][k];
    fs[u][1] += fs[to][k] + dp[to][k];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}

```