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 type {
  long long x, y, s;
} S;
int n, K, i, j, k, l;
long long s, ans;
bool operator<(type a, type b) { return a.s < b.s; }
priority_queue<type> hp;
long long js(long long n, long long k) {
  return (n / k) * n + (n % k) * (n / k + 1);
}
int main() {
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &s), ans += s * s;
    if (s > 1) hp.push({s, 1, s * s - js(s, 2)});
  }
  for (i = n + 1; i <= K; i++) {
    S = hp.top(), hp.pop();
    ans -= S.s;
    ++S.y;
    if (S.x > S.y) hp.push({S.x, S.y, js(S.x, S.y) - js(S.x, S.y + 1)});
  }
  printf("%lld\n", ans);
  fclose(stdin);
  fclose(stdout);
  return 0;
}
