#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200007, MAXK = 5;
vector<int> G[MAXN];
int n, k;
bool used[MAXN];
long long ans = 0, cnt[MAXN][MAXK], sum[MAXN][MAXK];
void dfs(int v, bool root) {
  used[v] = true;
  cnt[v][0]++;
  for (int i = (0); i < (((int)(G[v]).size())); ++i) {
    int to = G[v][i];
    if (!used[to]) {
      dfs(to, false);
      for (int j = (0); j < (k); ++j) {
        cnt[v][(j + 1) % k] += cnt[to][j];
        sum[v][(j + 1) % k] += sum[to][j] + cnt[to][j];
      }
    }
  }
  for (int r1 = (0); r1 < (k); ++r1)
    for (int r2 = (r1); r2 < (k); ++r2) {
      long long cnt1, cnt2, sum1, sum2;
      cnt1 = cnt[v][r1];
      cnt2 = cnt[v][r2];
      sum1 = sum[v][r1];
      sum2 = sum[v][r2];
      int delta = (r1 + r2) / k;
      if ((r1 + r2) % k != 0) ++delta;
      if (r1 != r2) {
        ans += (cnt1 * sum2 + cnt2 * sum1 - cnt1 * cnt2 * (r1 + r2)) / k;
        ans += cnt1 * cnt2 * delta;
      } else {
        ans += ((cnt1 - 1) * sum2 - cnt1 * (cnt2 - 1) * (r1 + r2) / 2) / k;
        ans += cnt1 * (cnt2 - 1) * delta / 2;
      }
      if (!root) {
        cnt1 = cnt[v][(r1 + k - 1) % k];
        cnt2 = cnt[v][(r2 + k - 1) % k];
        sum1 = sum[v][(r1 + k - 1) % k] + cnt1;
        sum2 = sum[v][(r2 + k - 1) % k] + cnt2;
        if (r1 != r2) {
          ans -= (cnt1 * sum2 + cnt2 * sum1 - cnt1 * cnt2 * (r1 + r2)) / k;
          ans -= cnt1 * cnt2 * delta;
        } else {
          ans -= ((cnt1 - 1) * sum2 - cnt1 * (cnt2 - 1) * (r1 + r2) / 2) / k;
          ans -= cnt1 * (cnt2 - 1) * delta / 2;
        }
      }
    }
}
int main() {
  cin >> n >> k;
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1);
    G[b - 1].push_back(a - 1);
  }
  memset(used, false, MAXN);
  dfs(0, true);
  cout << ans;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 MAXN = 200007, MAXK = 5;
vector<int> G[MAXN];
int n, k;
bool used[MAXN];
long long ans = 0, cnt[MAXN][MAXK], sum[MAXN][MAXK];
void dfs(int v, bool root) {
  used[v] = true;
  cnt[v][0]++;
  for (int i = (0); i < (((int)(G[v]).size())); ++i) {
    int to = G[v][i];
    if (!used[to]) {
      dfs(to, false);
      for (int j = (0); j < (k); ++j) {
        cnt[v][(j + 1) % k] += cnt[to][j];
        sum[v][(j + 1) % k] += sum[to][j] + cnt[to][j];
      }
    }
  }
  for (int r1 = (0); r1 < (k); ++r1)
    for (int r2 = (r1); r2 < (k); ++r2) {
      long long cnt1, cnt2, sum1, sum2;
      cnt1 = cnt[v][r1];
      cnt2 = cnt[v][r2];
      sum1 = sum[v][r1];
      sum2 = sum[v][r2];
      int delta = (r1 + r2) / k;
      if ((r1 + r2) % k != 0) ++delta;
      if (r1 != r2) {
        ans += (cnt1 * sum2 + cnt2 * sum1 - cnt1 * cnt2 * (r1 + r2)) / k;
        ans += cnt1 * cnt2 * delta;
      } else {
        ans += ((cnt1 - 1) * sum2 - cnt1 * (cnt2 - 1) * (r1 + r2) / 2) / k;
        ans += cnt1 * (cnt2 - 1) * delta / 2;
      }
      if (!root) {
        cnt1 = cnt[v][(r1 + k - 1) % k];
        cnt2 = cnt[v][(r2 + k - 1) % k];
        sum1 = sum[v][(r1 + k - 1) % k] + cnt1;
        sum2 = sum[v][(r2 + k - 1) % k] + cnt2;
        if (r1 != r2) {
          ans -= (cnt1 * sum2 + cnt2 * sum1 - cnt1 * cnt2 * (r1 + r2)) / k;
          ans -= cnt1 * cnt2 * delta;
        } else {
          ans -= ((cnt1 - 1) * sum2 - cnt1 * (cnt2 - 1) * (r1 + r2) / 2) / k;
          ans -= cnt1 * (cnt2 - 1) * delta / 2;
        }
      }
    }
}
int main() {
  cin >> n >> k;
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1);
    G[b - 1].push_back(a - 1);
  }
  memset(used, false, MAXN);
  dfs(0, true);
  cout << ans;
  return 0;
}

```