#include <bits/stdc++.h>
using namespace std;
int n, k;
long long sol = 0;
vector<vector<int> > G;
vector<vector<pair<long long, int> > > d;
void dfs(int u, int from) {
  d[u].resize(k);
  d[u][0].second += 1;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != from) {
      dfs(v, u);
      for (int l = 0; l < k; ++l) {
        for (int p = 0; p < k; ++p) {
          long long fv = d[v][l].first;
          int cntv = d[v][l].second;
          sol += fv * d[u][p].second + d[u][p].first * cntv;
          if (l == 0 && p == 0)
            sol += d[u][p].second * 1LL * cntv;
          else if (l + p + 1 <= k && l > 0 && p > 0)
            sol -= d[u][p].second * 1LL * cntv;
        }
      }
      d[u][1 % k].first += d[v][0].second;
      for (int j = 0; j < k; ++j) {
        d[u][j].first += d[v][(j - 1 + k) % k].first;
        d[u][j].second += d[v][(j - 1 + k) % k].second;
      }
    }
  }
}
int main() {
  cin >> n >> k;
  G.resize(n);
  d.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0, -1);
  cout << sol;
  return 0;
}

### 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;
int n, k;
long long sol = 0;
vector<vector<int> > G;
vector<vector<pair<long long, int> > > d;
void dfs(int u, int from) {
  d[u].resize(k);
  d[u][0].second += 1;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v != from) {
      dfs(v, u);
      for (int l = 0; l < k; ++l) {
        for (int p = 0; p < k; ++p) {
          long long fv = d[v][l].first;
          int cntv = d[v][l].second;
          sol += fv * d[u][p].second + d[u][p].first * cntv;
          if (l == 0 && p == 0)
            sol += d[u][p].second * 1LL * cntv;
          else if (l + p + 1 <= k && l > 0 && p > 0)
            sol -= d[u][p].second * 1LL * cntv;
        }
      }
      d[u][1 % k].first += d[v][0].second;
      for (int j = 0; j < k; ++j) {
        d[u][j].first += d[v][(j - 1 + k) % k].first;
        d[u][j].second += d[v][(j - 1 + k) % k].second;
      }
    }
  }
}
int main() {
  cin >> n >> k;
  G.resize(n);
  d.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0, -1);
  cout << sol;
  return 0;
}

```