#include <bits/stdc++.h>
using namespace std;
struct bkn {
  vector<int> son;
} f[200100];
long long k;
long long n, sum[200100][2], dp[200100][6][2];
void dfs(int x, int fa) {
  long long i;
  dp[x][0][0]++;
  for (i = 0; i < f[x].son.size(); i++) {
    long long y = f[x].son[i];
    if (y == fa) continue;
    dp[y][1][0] = dp[x][k][0] + dp[x][k][1];
    sum[y][0] = sum[x][0] + sum[x][1] + 1 + dp[x][k][1] + dp[x][k][0];
    for (int j = 0; j <= k - 1; j++) {
      dp[y][j + 1][0] += dp[x][j][1] + dp[x][j][0];
    }
    dfs(y, x);
    sum[x][1] += dp[y][k][1] + sum[y][1] + 1;
    dp[x][1][1] += dp[y][k][1];
    for (int j = 0; j <= k - 1; j++) {
      dp[x][j + 1][1] += dp[y][j][1];
    }
  }
  dp[x][0][1]++;
}
int main() {
  int x, y, i, j;
  while (cin >> n >> k) {
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    for (i = 1; i <= n - 1; i++) {
      cin >> x >> y;
      f[x].son.push_back(y);
      f[y].son.push_back(x);
    }
    dfs(1, 0);
    long long pp = 0;
    for (i = 1; i <= n; i++) {
      pp += sum[i][0];
    }
    cout << pp << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 bkn {
  vector<int> son;
} f[200100];
long long k;
long long n, sum[200100][2], dp[200100][6][2];
void dfs(int x, int fa) {
  long long i;
  dp[x][0][0]++;
  for (i = 0; i < f[x].son.size(); i++) {
    long long y = f[x].son[i];
    if (y == fa) continue;
    dp[y][1][0] = dp[x][k][0] + dp[x][k][1];
    sum[y][0] = sum[x][0] + sum[x][1] + 1 + dp[x][k][1] + dp[x][k][0];
    for (int j = 0; j <= k - 1; j++) {
      dp[y][j + 1][0] += dp[x][j][1] + dp[x][j][0];
    }
    dfs(y, x);
    sum[x][1] += dp[y][k][1] + sum[y][1] + 1;
    dp[x][1][1] += dp[y][k][1];
    for (int j = 0; j <= k - 1; j++) {
      dp[x][j + 1][1] += dp[y][j][1];
    }
  }
  dp[x][0][1]++;
}
int main() {
  int x, y, i, j;
  while (cin >> n >> k) {
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    for (i = 1; i <= n - 1; i++) {
      cin >> x >> y;
      f[x].son.push_back(y);
      f[y].son.push_back(x);
    }
    dfs(1, 0);
    long long pp = 0;
    for (i = 1; i <= n; i++) {
      pp += sum[i][0];
    }
    cout << pp << endl;
  }
}

```