#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > gr;
long long n, k;
long long ans = 0;
void dfs(long long v, long long p, vector<long long> &cnt,
         vector<long long> &dp) {
  for (long long u : gr[v]) {
    if (u == p) continue;
    vector<long long> cnt1(k, 0), dp1(k, 0);
    dfs(u, v, cnt1, dp1);
    vector<long long> dp2(k, 0), cnt2(k, 0);
    for (long long i = 0; i < k - 1; i++) {
      dp2[i + 1] = dp1[i];
      cnt2[i + 1] = cnt1[i];
    }
    dp2[0] = dp1[k - 1] + cnt1[k - 1];
    cnt2[0] = cnt1[k - 1];
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        if (i + j == 0) {
          ans += dp[i] * cnt2[j] + dp2[j] * cnt[i];
        } else if (i + j > k) {
          ans += dp[i] * cnt2[j] + dp2[j] * cnt[i] + 2 * cnt[i] * cnt2[j];
        } else {
          ans += cnt[i] * cnt2[j] + dp[i] * cnt2[j] + dp2[j] * cnt[i];
        }
      }
    }
    for (long long i = 0; i < k; i++) {
      ans += dp2[i];
      if (i != 0) ans += cnt2[i];
      dp[i] += dp2[i];
      cnt[i] += cnt2[i];
    }
  }
  cnt[0]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  gr.resize(n);
  for (long long i = 1; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    gr[l].push_back(r);
    gr[r].push_back(l);
  }
  vector<long long> cnt(k, 0), dp(k, 0);
  dfs(0, -1, cnt, dp);
  cout << ans;
}

### 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;
vector<vector<long long> > gr;
long long n, k;
long long ans = 0;
void dfs(long long v, long long p, vector<long long> &cnt,
         vector<long long> &dp) {
  for (long long u : gr[v]) {
    if (u == p) continue;
    vector<long long> cnt1(k, 0), dp1(k, 0);
    dfs(u, v, cnt1, dp1);
    vector<long long> dp2(k, 0), cnt2(k, 0);
    for (long long i = 0; i < k - 1; i++) {
      dp2[i + 1] = dp1[i];
      cnt2[i + 1] = cnt1[i];
    }
    dp2[0] = dp1[k - 1] + cnt1[k - 1];
    cnt2[0] = cnt1[k - 1];
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        if (i + j == 0) {
          ans += dp[i] * cnt2[j] + dp2[j] * cnt[i];
        } else if (i + j > k) {
          ans += dp[i] * cnt2[j] + dp2[j] * cnt[i] + 2 * cnt[i] * cnt2[j];
        } else {
          ans += cnt[i] * cnt2[j] + dp[i] * cnt2[j] + dp2[j] * cnt[i];
        }
      }
    }
    for (long long i = 0; i < k; i++) {
      ans += dp2[i];
      if (i != 0) ans += cnt2[i];
      dp[i] += dp2[i];
      cnt[i] += cnt2[i];
    }
  }
  cnt[0]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  gr.resize(n);
  for (long long i = 1; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    gr[l].push_back(r);
    gr[r].push_back(l);
  }
  vector<long long> cnt(k, 0), dp(k, 0);
  dfs(0, -1, cnt, dp);
  cout << ans;
}

```