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 INF = 1000 * 1000 * 1000;
const int MOD = 1000 * 1000 * 1000 + 7;
const long INF_L = 1000 * 1000 * 1000;
const long long INF_LL = 1LL * INF_L * INF_L;
const long long MOD_LL = 1000000000007LL;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
long long pow(long long n, long long a, long long mod) {
  long long res = 1;
  while (a) {
    if (a & 1) res = (res * n) % mod;
    n = (n * n) % mod;
    a >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
const int N = 200 * 1000 + 10;
const int M = 4000;
const int MAX_K = 5;
pair<long long, long long> d[N][MAX_K];
long long ans;
void dfs(int v, int p, const vector<vector<int>> &t, int k) {
  vector<pair<long long, long long>> tmp(k, make_pair(0, 0));
  for (int i = 0; i < t[v].size(); ++i) {
    int son = t[v][i];
    if (p != son) {
      dfs(son, v, t, k);
      for (int j1 = 0; j1 < k; ++j1) {
        for (int j2 = 0; j2 < k; ++j2) {
          ans +=
              tmp[j1].first * (d[son][j2].second) +
              tmp[j1].second * (d[son][j2].first) +
              ((j1 + j2 + 1 + k - 1) / k) * (tmp[j1].first * d[son][j2].first);
        }
        ans += tmp[j1].second + ((j1 + 1 + k - 1) / k) * (tmp[j1].first);
      }
      for (int j = 0; j < k; ++j) {
        d[v][(j + 1) % k].first += d[son][j].first;
        d[v][(j + 1) % k].second +=
            d[son][j].second + (j + 1 >= k ? d[son][j].first : 0);
        tmp[(j + 1) % k].first += d[son][j].first;
        tmp[(j + 1) % k].second +=
            d[son][j].second + (j + 1 >= k ? d[son][j].first : 0);
      }
      d[v][1 % k].first += 1;
      tmp[1 % k].first += 1;
      if (k == 1) {
        d[v][0].second += 1;
        tmp[0].second += 1;
      }
    }
  }
  for (int i = 0; i < k; ++i) {
    ans += (d[v][i].second + (i ? 1 : 0) * d[v][i].first);
  }
}
int main() {
  int n, k, u, v;
  cin >> n >> k;
  vector<vector<int>> t(n);
  for (int i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    --u;
    --v;
    t[u].emplace_back(v);
    t[v].emplace_back(u);
  }
  dfs(0, -1, t, k);
  cout << ans << endl;
  return 0;
}
