#include <bits/stdc++.h>
using namespace std;
struct mydfs {
  int k;
  vector<vector<int>> *padj;
  array<array<int, 5>, 5> precomputedAdder;
  vector<int> subtreeSize;
  vector<vector<int>> subtreeRemainders;
  uint64_t ktimesans;
  int maxnode;
  void init(vector<vector<int>> &adj, int maxnodein, int kin) {
    maxnode = maxnodein;
    k = kin;
    ktimesans = 0;
    padj = &adj;
    subtreeSize.clear();
    subtreeSize.resize(maxnode + 1, 0);
    subtreeRemainders.clear();
    subtreeRemainders.resize(maxnode + 1);
    for (int i = 0; i <= maxnode; i++) {
      subtreeRemainders[i].resize(5, 0);
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        precomputedAdder[i][j] =
            ((i + j + 1) % k == 0) ? 0 : k - ((i + j + 1) % k);
      }
    }
  }
  uint64_t dfs(int n, int p) {
    for (auto nn : (*padj)[n]) {
      if (nn == p) continue;
      dfs(nn, n);
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          ktimesans += (uint64_t)precomputedAdder[i][j] *
                       (uint64_t)subtreeRemainders[n][i] *
                       (uint64_t)subtreeRemainders[nn][j];
        }
      }
      for (int j = 0; j < k; j++) {
        ktimesans += (uint64_t)precomputedAdder[0][j] *
                     (uint64_t)subtreeRemainders[nn][j];
      }
      for (int j = 0; j < k; j++) {
        if (j + 1 == k) {
          subtreeRemainders[n][0] += subtreeRemainders[nn][j];
        } else {
          subtreeRemainders[n][j + 1] += subtreeRemainders[nn][j];
        }
      }
      subtreeSize[n] += subtreeSize[nn];
    }
    subtreeSize[n] += 1;
    subtreeRemainders[n][0] += 1;
    ktimesans +=
        (uint64_t)subtreeSize[n] * (uint64_t)(maxnode - subtreeSize[n]);
    return ktimesans;
  }
};
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> adj(n + 1);
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  mydfs d;
  d.init(adj, n, k);
  uint64_t ans = d.dfs(1, -1);
  assert(ans % k == 0);
  cout << ans / k << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
struct mydfs {
  int k;
  vector<vector<int>> *padj;
  array<array<int, 5>, 5> precomputedAdder;
  vector<int> subtreeSize;
  vector<vector<int>> subtreeRemainders;
  uint64_t ktimesans;
  int maxnode;
  void init(vector<vector<int>> &adj, int maxnodein, int kin) {
    maxnode = maxnodein;
    k = kin;
    ktimesans = 0;
    padj = &adj;
    subtreeSize.clear();
    subtreeSize.resize(maxnode + 1, 0);
    subtreeRemainders.clear();
    subtreeRemainders.resize(maxnode + 1);
    for (int i = 0; i <= maxnode; i++) {
      subtreeRemainders[i].resize(5, 0);
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        precomputedAdder[i][j] =
            ((i + j + 1) % k == 0) ? 0 : k - ((i + j + 1) % k);
      }
    }
  }
  uint64_t dfs(int n, int p) {
    for (auto nn : (*padj)[n]) {
      if (nn == p) continue;
      dfs(nn, n);
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          ktimesans += (uint64_t)precomputedAdder[i][j] *
                       (uint64_t)subtreeRemainders[n][i] *
                       (uint64_t)subtreeRemainders[nn][j];
        }
      }
      for (int j = 0; j < k; j++) {
        ktimesans += (uint64_t)precomputedAdder[0][j] *
                     (uint64_t)subtreeRemainders[nn][j];
      }
      for (int j = 0; j < k; j++) {
        if (j + 1 == k) {
          subtreeRemainders[n][0] += subtreeRemainders[nn][j];
        } else {
          subtreeRemainders[n][j + 1] += subtreeRemainders[nn][j];
        }
      }
      subtreeSize[n] += subtreeSize[nn];
    }
    subtreeSize[n] += 1;
    subtreeRemainders[n][0] += 1;
    ktimesans +=
        (uint64_t)subtreeSize[n] * (uint64_t)(maxnode - subtreeSize[n]);
    return ktimesans;
  }
};
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> adj(n + 1);
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  mydfs d;
  d.init(adj, n, k);
  uint64_t ans = d.dfs(1, -1);
  assert(ans % k == 0);
  cout << ans / k << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

```