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.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int maxn = 2e5 + 1;
int n, k;
vector<vector<int>> tree(maxn);
vector<vector<long long>> dp(maxn, vector<long long>(5)),
    dp2(maxn, vector<long long>(5)), dp3(maxn, vector<long long>(5)),
    dp4(maxn, vector<long long>(5));
void dfs(int x, int p = -1) {
  dp2[x][0]++;
  for (int y : tree[x]) {
    if (y == p) continue;
    dfs(y, x);
  }
  if (p != -1) {
    if (k == 1) {
      dp2[p][0] += dp2[x][0];
      dp[p][0] += dp[x][0] + dp2[x][0];
    } else {
      for (int i = 1; i <= k - 1; i++) {
        dp[p][(i + 1) % k] += dp[x][i];
        dp2[p][(i + 1) % k] += dp2[x][i];
      }
      dp[p][1] += dp[x][0] + dp2[x][0];
      dp2[p][1] += dp2[x][0];
    }
  }
}
void dfs2(int x, int p = -1) {
  for (int y : tree[x]) {
    if (y == p) continue;
    if (k == 1) {
      dp3[y][0] = dp3[x][0] + n - 2 * dp2[y][0];
    } else {
      for (int i = 1; i <= k - 1; i++) {
        int j = (i + 1) % k;
        dp3[y][j] =
            dp3[x][i] - dp[y][i - 1] + dp[y][j] - (i == 1 ? dp2[y][i - 1] : 0);
        dp4[y][j] = dp4[x][i] - dp2[y][i - 1] + dp2[y][j];
      }
      dp3[y][1] =
          dp3[x][0] + dp4[x][0] - dp2[y][k - 1] - dp[y][k - 1] + dp[y][1];
      dp4[y][1] = dp4[x][0] - dp2[y][k - 1] + dp2[y][1];
    }
    dfs2(y, x);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    tree[x].push_back(y);
    tree[y].push_back(x);
  }
  dfs(1);
  for (int i = 0; i < k; i++) {
    dp3[1][i] = dp[1][i];
    dp4[1][i] = dp2[1][i];
  }
  dfs2(1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      ans += dp3[i][j];
    }
  }
  cout << ans / 2 << '\n';
  return 0;
}
