#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 200010;
const int K = 5;
int n, k;
vector<int> adj[N];
ll ans = 0;
vector<pair<ll, int>> dfs(int u = 1, int p = -1) {
  vector<pair<ll, int>> ret(k);
  vector<vector<pair<ll, int>>> children;
  for (int v : adj[u]) {
    if (v == p) continue;
    children.emplace_back(dfs(v, u));
    auto& vec = children.back();
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first += vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second += vec[i].second;
    }
  }
  ll tmp = 0;
  for (auto& vec : children) {
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first -= vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second -= vec[i].second;
    }
    for (int i = 0; i < k; i++) {
      ll d = vec[i].first + (i == 0) * vec[i].second;
      ll q = vec[i].second;
      for (int j = 0; j < k; j++) {
        tmp += d * ret[j].second + ret[j].first * q;
        int fr = ((i + 1) % k > 0) + (j > 0);
        int to = (((i + 1) % k) + j + k - 1) / k;
        tmp += q * ret[j].second * (to - fr);
      }
    }
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first += vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second += vec[i].second;
    }
  }
  tmp /= 2;
  for (int j = 0; j < k; j++) {
    tmp += ret[j].first;
  }
  ret[0].second++;
  ans += tmp;
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  printf("%lld\n", ans);
  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;
using ll = long long;
const int N = 200010;
const int K = 5;
int n, k;
vector<int> adj[N];
ll ans = 0;
vector<pair<ll, int>> dfs(int u = 1, int p = -1) {
  vector<pair<ll, int>> ret(k);
  vector<vector<pair<ll, int>>> children;
  for (int v : adj[u]) {
    if (v == p) continue;
    children.emplace_back(dfs(v, u));
    auto& vec = children.back();
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first += vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second += vec[i].second;
    }
  }
  ll tmp = 0;
  for (auto& vec : children) {
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first -= vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second -= vec[i].second;
    }
    for (int i = 0; i < k; i++) {
      ll d = vec[i].first + (i == 0) * vec[i].second;
      ll q = vec[i].second;
      for (int j = 0; j < k; j++) {
        tmp += d * ret[j].second + ret[j].first * q;
        int fr = ((i + 1) % k > 0) + (j > 0);
        int to = (((i + 1) % k) + j + k - 1) / k;
        tmp += q * ret[j].second * (to - fr);
      }
    }
    for (int i = 0; i < k; i++) {
      ret[(i + 1) % k].first += vec[i].first + (i == 0) * vec[i].second;
      ret[(i + 1) % k].second += vec[i].second;
    }
  }
  tmp /= 2;
  for (int j = 0; j < k; j++) {
    tmp += ret[j].first;
  }
  ret[0].second++;
  ans += tmp;
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  printf("%lld\n", ans);
  return 0;
}

```