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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = x * z;
    x = x * x;
    y /= 2;
  }
  return z;
}
pair<int, int> func(int x, int div) {
  if (x % div == 0) return make_pair(x / div, div);
  int val = x / div + 1;
  int val2 = x - (x / div) * div;
  return make_pair(val, val2);
}
long long cost(int a, int b) {
  long long ans = 0;
  pair<int, int> p = func(a, b);
  ans += p.second * binpow(p.first, 2);
  ans += (b - p.second) * binpow(p.first - 1, 2);
  return ans;
}
int main() {
  int n, k;
  cin >> n >> k;
  int arr[n];
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  int cuts[n];
  for (int i = 0; i < n; i++) cuts[i] = 1;
  priority_queue<pair<long long, int>> pq;
  for (int i = 0; i < n; i++)
    pq.push(make_pair(cost(arr[i], cuts[i]) - cost(arr[i], cuts[i] + 1), i));
  for (int i = 0; i < k - n; i++) {
    int id = pq.top().second;
    pq.pop();
    cuts[id]++;
    pq.push(
        make_pair(cost(arr[id], cuts[id]) - cost(arr[id], cuts[id] + 1), id));
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    pair<int, int> p = func(arr[i], cuts[i]);
    ans += p.second * binpow(p.first, 2);
    ans += (cuts[i] - p.second) * binpow(p.first - 1, 2);
  }
  cout << ans << endl;
  return 0;
}
