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;
struct aaa {
  long long x, y;
};
priority_queue<aaa> q;
bool operator<(aaa a, aaa b) { return a.x < b.x; }
long long n, k, ans, a[200010], mn, b[200010];
long long pd(long long mid) {
  long long i, xlh = 0;
  for (i = 1; i <= n; i++) xlh += (a[i] - 1) / mid + 1;
  return xlh <= k;
}
long long get(long long x, long long y) {
  long long w = x / y, yy = x % y, xx = y - yy;
  return xx * w * w + yy * (w + 1) * (w + 1);
}
int main() {
  long long i, j, hh, w, x, y;
  aaa now;
  scanf("%lld%lld", &n, &k);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 1; i <= n; i++) {
    ans += a[i] * a[i];
    b[i] = 1;
    if (a[i] > 1) q.push(aaa{get(a[i], 1) - get(a[i], 2), i});
  }
  k -= n;
  while (k > 0) {
    k--;
    now = q.top();
    q.pop();
    ans -= now.x;
    b[now.y]++;
    if (a[now.y] > b[now.y])
      q.push(aaa{get(a[now.y], b[now.y]) - get(a[now.y], b[now.y] + 1), now.y});
  }
  printf("%lld", ans);
}
