#include <bits/stdc++.h>
using namespace std;
long long int dp[200005][5];
long long int subtree[200005];
vector<long long int> v[200005];
long long int N, K;
long long int ans = 0;
void dfs(long long int u, long long int par) {
  dp[u][0] = 1;
  subtree[u] = 1;
  for (long long int i = 0; i < v[u].size(); i++) {
    if (v[u][i] != par) {
      dfs(v[u][i], u);
      for (long long int j = 0; j < 5; j++) {
        for (long long int k = 0; k < 5; k++) {
          ans += ((K - (j + k + 1) % K) % K) * dp[u][j] * dp[v[u][i]][k];
        }
      }
      for (long long int j = 0; j < 5; j++)
        dp[u][(j + 1) % K] += dp[v[u][i]][j];
      subtree[u] += subtree[v[u][i]];
    }
  }
  ans += (subtree[u]) * (N - subtree[u]);
}
int main() {
  long long int i, j, k;
  cin >> N >> K;
  for (i = 0; i < N - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  cout << ans / K << 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;
long long int dp[200005][5];
long long int subtree[200005];
vector<long long int> v[200005];
long long int N, K;
long long int ans = 0;
void dfs(long long int u, long long int par) {
  dp[u][0] = 1;
  subtree[u] = 1;
  for (long long int i = 0; i < v[u].size(); i++) {
    if (v[u][i] != par) {
      dfs(v[u][i], u);
      for (long long int j = 0; j < 5; j++) {
        for (long long int k = 0; k < 5; k++) {
          ans += ((K - (j + k + 1) % K) % K) * dp[u][j] * dp[v[u][i]][k];
        }
      }
      for (long long int j = 0; j < 5; j++)
        dp[u][(j + 1) % K] += dp[v[u][i]][j];
      subtree[u] += subtree[v[u][i]];
    }
  }
  ans += (subtree[u]) * (N - subtree[u]);
}
int main() {
  long long int i, j, k;
  cin >> N >> K;
  for (i = 0; i < N - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  cout << ans / K << endl;
  return 0;
}

```