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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int arr[100009], val[100009], n, k;
int ok(int x) {
  long long res = 0;
  for (int i = 1; i <= n; i++) res += (arr[i] + x - 1) / x;
  return (res <= k);
}
long long f(int x, int y) {
  int a = x % y, aa = x / y + 1;
  int b = y - a, bb = x / y;
  return aa * 1LL * aa * a + bb * 1LL * bb * b;
}
set<pair<long long, int> > s;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    val[i] = 1;
    if (arr[i] > 1)
      s.insert(make_pair(f(arr[i], val[i] + 1) - f(arr[i], val[i]), i));
  }
  for (int j = 0; j < k - n; j++) {
    pair<long long, int> nd = *s.begin();
    s.erase(nd);
    int i = nd.second;
    val[i]++;
    if (val[i] < arr[i])
      s.insert(make_pair(f(arr[i], val[i] + 1) - f(arr[i], val[i]), i));
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) res += f(arr[i], val[i]);
  printf("%lld\n", res);
  return 0;
}
