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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long mat[200005][5], dp[200005];
vector<long long> adj[200005];
long long k;
void dfs(long long n = 1, long long par = 0) {
  for (auto &cv : adj[n]) {
    if (cv == par) continue;
    dfs(cv, n);
    dp[n] += dp[cv] + mat[cv][0];
    for (long long i = 0; i < k; i++) {
      mat[n][(i + 1) % k] += mat[cv][i];
    }
  }
  mat[n][0]++;
}
void dfs1(long long n = 1, long long par = 0) {
  long long pre = dp[n] + mat[n][0];
  pre = dp[par] - pre;
  if (par) dp[n] += pre + mat[par][0] - mat[n][k - 1];
  long long ad[5];
  for (long long i = 0; i < k; i++) {
    if (par) {
      ad[(i + 1) % k] = mat[par][i] - mat[n][(i - 1 + k) % k];
    }
  }
  for (long long i = 0; i < k; i++) {
    if (par) mat[n][i] += ad[i];
  }
  for (auto &cv : adj[n]) {
    if (cv == par) continue;
    dfs1(cv, n);
  }
}
void solve() {
  long long i, j, l, r, m, n, a, b, c, d, x, y, z, t;
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs();
  dfs1();
  long long ans = 0;
  for (i = 1; i <= n; i++) ans += dp[i];
  cout << ans / 2 << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
