#include <bits/stdc++.h>
using namespace std;
const int mxn = 2e5 + 10;
const int mxlg = 14;
const int mod = 1e9 + 7;
vector<int> adj[mxn];
long long sz[mxn], dp[mxn][5], n, k;
long long ans;
void dfs(int node = 1, int parent = -1) {
  sz[node] = 1;
  for (auto x : adj[node])
    if (x != parent) {
      dfs(x, node);
      sz[node] += sz[x];
    }
  vector<long long> tmp(k, 0);
  for (auto x : adj[node])
    if (x != parent)
      for (int i = 0; i < k; i++) {
        tmp[(i + 1) % k] += dp[x][i];
      }
  long long tempAnswer = 0;
  for (auto x : adj[node])
    if (x != parent) {
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          tempAnswer += ((3 * k - 1 - i - j) % k) * dp[x][i] *
                        (tmp[j] - dp[x][(j + k - 1) % k]);
        }
    }
  tempAnswer /= 2;
  for (int i = 1; i < k; i++) tempAnswer += (k - i) * tmp[i];
  ans += tempAnswer;
  for (int i = 0; i < k; i++) dp[node][i] = tmp[i];
  dp[node][0]++;
}
void dfs2(int node = 1, int parent = -1) {
  for (auto x : adj[node])
    if (x != parent) {
      dfs2(x, node);
      ans += 1ll * sz[x] * (n - sz[x]);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  dfs2();
  assert(ans % k == 0);
  cout << ans / k << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
const int mxn = 2e5 + 10;
const int mxlg = 14;
const int mod = 1e9 + 7;
vector<int> adj[mxn];
long long sz[mxn], dp[mxn][5], n, k;
long long ans;
void dfs(int node = 1, int parent = -1) {
  sz[node] = 1;
  for (auto x : adj[node])
    if (x != parent) {
      dfs(x, node);
      sz[node] += sz[x];
    }
  vector<long long> tmp(k, 0);
  for (auto x : adj[node])
    if (x != parent)
      for (int i = 0; i < k; i++) {
        tmp[(i + 1) % k] += dp[x][i];
      }
  long long tempAnswer = 0;
  for (auto x : adj[node])
    if (x != parent) {
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          tempAnswer += ((3 * k - 1 - i - j) % k) * dp[x][i] *
                        (tmp[j] - dp[x][(j + k - 1) % k]);
        }
    }
  tempAnswer /= 2;
  for (int i = 1; i < k; i++) tempAnswer += (k - i) * tmp[i];
  ans += tempAnswer;
  for (int i = 0; i < k; i++) dp[node][i] = tmp[i];
  dp[node][0]++;
}
void dfs2(int node = 1, int parent = -1) {
  for (auto x : adj[node])
    if (x != parent) {
      dfs2(x, node);
      ans += 1ll * sz[x] * (n - sz[x]);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  dfs2();
  assert(ans % k == 0);
  cout << ans / k << '\n';
  return 0;
}

```