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;
const int N = 200005;
struct node {
  long long val;
  int num, id;
  bool operator<(const node &a) const { return a.val > val; }
};
long long solve(long long x, int num) {
  long long ret = x / num, res = x % num;
  long long ans =
      1ll * res * (ret + 1) * (ret + 1) + 1ll * (num - res) * ret * ret;
  return ans;
}
priority_queue<node> q;
int a[N];
int n, k;
long long sum = 0;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    sum += (1ll * a[i] * a[i]);
    if (a[i] == 1) continue;
    node nw;
    nw.val = 1ll * a[i] * a[i] - solve(a[i], 2);
    nw.num = 2;
    nw.id = i;
    q.push(nw);
  }
  if (n == k) {
    printf("%lld\n", sum);
    return 0;
  }
  int cnt = 0;
  while (!q.empty()) {
    ++cnt;
    node st = q.top();
    q.pop();
    sum -= st.val;
    if (cnt == (k - n)) {
      printf("%lld\n", sum);
      return 0;
    }
    if (a[st.id] < (st.num + 1)) continue;
    node nw;
    nw.val = solve(1ll * a[st.id], st.num) - solve(1ll * a[st.id], st.num + 1);
    nw.num = st.num + 1;
    nw.id = st.id;
    q.push(nw);
  }
  return 0;
}
