There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.

Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.

Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.

Input

The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.

It is guaranteed that the sum of a_i is at least k.

Output

Output one integer: the minimum sum of time taken for rabbits to eat carrots.

Examples

Input


3 6
5 3 1


Output


15


Input


1 4
19


Output


91

Note

For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15

For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
#include <bits/stdc++.h>
using namespace std;
string getString(char x) {
  string s(1, x);
  return s;
}
int highestPowerof2(long long int n) {
  if (n == 0) return 0;
  int p = (int)log2(n);
  return p;
}
long long int calc(int num, int div) {
  int p = num / div;
  int k = num - p * div;
  long long int ans = 0;
  ans += ((long long int)(p + 1) * (p + 1) * k);
  ans += ((long long int)(p)*p * (div - k));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    int n, k;
    cin >> n >> k;
    priority_queue<pair<long long int, int>> pq;
    vector<int> vec(n);
    vector<int> curr(n, 1);
    for (int i = 0; i < n; i++) {
      cin >> vec[i];
      long long int val = calc(vec[i], curr[i]) - calc(vec[i], curr[i] + 1);
      pq.push(pair<long long int, int>(val, i));
    }
    int req = k - n;
    while (req > 0) {
      req--;
      pair<long long int, int> p = pq.top();
      pq.pop();
      curr[p.second]++;
      int i = p.second;
      long long int val = calc(vec[i], curr[i]) - calc(vec[i], curr[i] + 1);
      p.first = val;
      pq.push(p);
    }
    long long int ans = 0;
    for (int i = 0; i < n; i++) {
      ans += calc(vec[i], curr[i]);
    }
    cout << ans << endl;
  }
  return 0;
}
