#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
const int N = 1e6 + 100;
int n, k;
vector<int> g[N];
int cnt[N][5];
long long totalSum[N][5];
long long ans[N];
void dfs(int v, int p = -1) {
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
  static int curCnt[5];
  static long long curSum[5];
  fill_n(curCnt, 5, 0);
  fill_n(curSum, 5, 0);
  curCnt[0]++;
  for (int to : g[v]) {
    if (p == to) {
      continue;
    }
    static int nextCnt[5];
    static long long nextTotalSum[5];
    fill_n(nextCnt, 5, 0);
    fill_n(nextTotalSum, 5, 0);
    for (int i = 0; i < k; i++) {
      nextCnt[(i + 1) % k] = cnt[to][i];
      nextTotalSum[(i + 1) % k] =
          totalSum[to][i] + 1ll * cnt[to][i] * ((i + 1) / k);
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        ans[v] += 1ll * curSum[i] * nextCnt[j] +
                  1ll * curCnt[i] * nextTotalSum[j] +
                  1ll * curCnt[i] * nextCnt[j] * ((i + j + k - 1) / k);
      }
    }
    for (int i = 0; i < k; i++) {
      curCnt[i] += nextCnt[i];
      curSum[i] += nextTotalSum[i];
    }
  }
  memcpy(cnt[v], curCnt, sizeof(curCnt));
  memcpy(totalSum[v], curSum, sizeof(curSum));
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v), u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += ::ans[i];
  }
  cout << ans << endl;
  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;
const int inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
const int N = 1e6 + 100;
int n, k;
vector<int> g[N];
int cnt[N][5];
long long totalSum[N][5];
long long ans[N];
void dfs(int v, int p = -1) {
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
  static int curCnt[5];
  static long long curSum[5];
  fill_n(curCnt, 5, 0);
  fill_n(curSum, 5, 0);
  curCnt[0]++;
  for (int to : g[v]) {
    if (p == to) {
      continue;
    }
    static int nextCnt[5];
    static long long nextTotalSum[5];
    fill_n(nextCnt, 5, 0);
    fill_n(nextTotalSum, 5, 0);
    for (int i = 0; i < k; i++) {
      nextCnt[(i + 1) % k] = cnt[to][i];
      nextTotalSum[(i + 1) % k] =
          totalSum[to][i] + 1ll * cnt[to][i] * ((i + 1) / k);
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        ans[v] += 1ll * curSum[i] * nextCnt[j] +
                  1ll * curCnt[i] * nextTotalSum[j] +
                  1ll * curCnt[i] * nextCnt[j] * ((i + j + k - 1) / k);
      }
    }
    for (int i = 0; i < k; i++) {
      curCnt[i] += nextCnt[i];
      curSum[i] += nextTotalSum[i];
    }
  }
  memcpy(cnt[v], curCnt, sizeof(curCnt));
  memcpy(totalSum[v], curSum, sizeof(curSum));
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v), u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += ::ans[i];
  }
  cout << ans << endl;
  return 0;
}

```