#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int K = 10;
int n, k;
vector<int> e[N];
long long d[N], siz[N], sum[N], num[N][K], f[N][K];
void dfs1(int u, int fa) {
  siz[u] = 1;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs1(v, u);
      siz[u] += siz[v];
    }
  }
}
void dfs2(int u, int fa) {
  d[u] = 0;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs2(v, u);
      sum[u] += (d[v] + siz[v]) * (siz[u] - siz[v]);
      d[u] += d[v] + siz[v];
    }
  }
}
void dfs3(int u, int fa) {
  for (int o = 0; o < k; ++o) num[u][o] = 0;
  num[u][0] = 1;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs3(v, u);
      for (int ou = 0; ou < k; ++ou) {
        for (int ov = 0; ov < k; ++ov) {
          int nw = (ou + ov + 1) % k;
          if (nw != 0) nw = k - nw;
          f[u][nw] += num[u][ou] * num[v][ov];
        }
      }
      for (int ov = 0; ov < k; ++ov) {
        num[u][(ov + 1) % k] += num[v][ov];
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  long long s = 0;
  for (int i = 1; i <= n; ++i) {
    s += sum[i];
  }
  for (int i = 1; i <= n; ++i) {
    for (int o = 0; o < k; ++o) {
      s += f[i][o] * o;
    }
  }
  cout << s / k << "\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 200005;
const int K = 10;
int n, k;
vector<int> e[N];
long long d[N], siz[N], sum[N], num[N][K], f[N][K];
void dfs1(int u, int fa) {
  siz[u] = 1;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs1(v, u);
      siz[u] += siz[v];
    }
  }
}
void dfs2(int u, int fa) {
  d[u] = 0;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs2(v, u);
      sum[u] += (d[v] + siz[v]) * (siz[u] - siz[v]);
      d[u] += d[v] + siz[v];
    }
  }
}
void dfs3(int u, int fa) {
  for (int o = 0; o < k; ++o) num[u][o] = 0;
  num[u][0] = 1;
  for (auto v : e[u]) {
    if (v != fa) {
      dfs3(v, u);
      for (int ou = 0; ou < k; ++ou) {
        for (int ov = 0; ov < k; ++ov) {
          int nw = (ou + ov + 1) % k;
          if (nw != 0) nw = k - nw;
          f[u][nw] += num[u][ou] * num[v][ov];
        }
      }
      for (int ov = 0; ov < k; ++ov) {
        num[u][(ov + 1) % k] += num[v][ov];
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  long long s = 0;
  for (int i = 1; i <= n; ++i) {
    s += sum[i];
  }
  for (int i = 1; i <= n; ++i) {
    for (int o = 0; o < k; ++o) {
      s += f[i][o] * o;
    }
  }
  cout << s / k << "\n";
  return 0;
}

```