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 = 998244353;
const int N = 200005;
const double PI = 4 * atan(1);
const double eps = 1e-7;
const long long oo = 1e10;
const int K = 205;
long long dp[N][5];
long long num[N][5];
long long n, k;
vector<long long> adj[N];
long long ans = 0;
void dfs(long long src, long long par) {
  num[src][0]++;
  for (auto t : adj[src]) {
    if (t == par) continue;
    dfs(t, src);
    for (int i = 0; i < k; i++) {
      long long x = dp[t][(i + k - 1) % k];
      if (i == 0) x += num[t][(i + k - 1) % k] * k;
      long long y = num[t][(i + k - 1) % k];
      for (int j = 0; j < k; j++) {
        ans += ((x / k) * num[src][j] + (dp[src][j] / k) * y);
        if (i + j > k)
          ans += 2 * (num[src][j] * y);
        else if (i + j > 0)
          ans += (num[src][j] * y);
      }
    }
    for (int i = 0; i < k; i++) {
      dp[src][(i + 1) % k] += (dp[t][i]);
      if (i == k - 1) dp[src][(i + 1) % k] += (num[t][i] * k);
      num[src][(i + 1) % k] += num[t][i];
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, 0);
  cout << ans;
  return 0;
}
