#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;
vector<int> g[1000001];
long long d[1000001], sum, dc[1000001][5], sz[1000001], cnt[1000001], cm[5], k;
int n;
void add(long long& a, long long b) { a += b; }
int dfs(int x, int p) {
  sz[x] = 1;
  dc[x][0] = 1;
  cnt[x] = 1;
  for (int to : g[x]) {
    if (to != p) {
      sz[x] += dfs(to, x);
      add(d[x], d[to] + cnt[to]);
      add(cnt[x], cnt[to]);
      for (int j = 0; j < k; j++) {
        add(dc[x][j], dc[to][j ? j - 1 : k - 1]);
      }
    }
  }
  long long pf[5] = {1, 0, 0, 0, 0};
  for (int to : g[x]) {
    if (to != p) {
      add(sum, (cnt[x] - cnt[to]) * (d[to] + cnt[to]));
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          long long s = pf[a] * dc[to][b ? b - 1 : k - 1];
          add(cm[(a + b) % k], s);
        }
        add(pf[a], dc[to][a ? a - 1 : k - 1]);
      }
    }
  }
  return sz[x];
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  for (int i = 1; i < k; i++) {
    add(sum, (k - i) * cm[i]);
  }
  cout << sum / k;
}

### Prompt

Please create a solution in cpp 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 long long INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;
vector<int> g[1000001];
long long d[1000001], sum, dc[1000001][5], sz[1000001], cnt[1000001], cm[5], k;
int n;
void add(long long& a, long long b) { a += b; }
int dfs(int x, int p) {
  sz[x] = 1;
  dc[x][0] = 1;
  cnt[x] = 1;
  for (int to : g[x]) {
    if (to != p) {
      sz[x] += dfs(to, x);
      add(d[x], d[to] + cnt[to]);
      add(cnt[x], cnt[to]);
      for (int j = 0; j < k; j++) {
        add(dc[x][j], dc[to][j ? j - 1 : k - 1]);
      }
    }
  }
  long long pf[5] = {1, 0, 0, 0, 0};
  for (int to : g[x]) {
    if (to != p) {
      add(sum, (cnt[x] - cnt[to]) * (d[to] + cnt[to]));
      for (int a = 0; a < k; a++) {
        for (int b = 0; b < k; b++) {
          long long s = pf[a] * dc[to][b ? b - 1 : k - 1];
          add(cm[(a + b) % k], s);
        }
        add(pf[a], dc[to][a ? a - 1 : k - 1]);
      }
    }
  }
  return sz[x];
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  for (int i = 1; i < k; i++) {
    add(sum, (k - i) * cm[i]);
  }
  cout << sum / k;
}

```