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 nmax = 1e6 + 42;
int n, wanted, inp[nmax];
long long eval(int i, int cur) {
  long long s = ((-inp[i]) % cur + cur) % cur;
  long long t = cur - s;
  long long d = (inp[i] + s) / cur;
  return 1LL * s * (d - 1) * (d - 1) + 1LL * t * d * d;
}
priority_queue<pair<long long, int> > pq;
int groups[nmax];
int main() {
  scanf("%i%i", &n, &wanted);
  wanted = wanted - n;
  for (int i = 1; i <= n; i++) scanf("%i", &inp[i]);
  long long outp = 0;
  for (int i = 1; i <= n; i++) {
    outp += 1LL * inp[i] * inp[i];
    groups[i]++;
    pq.push({eval(i, 1) - eval(i, 2), i});
  }
  while (wanted) {
    pair<long long, int> tp = pq.top();
    pq.pop();
    outp = outp - tp.first;
    groups[tp.second]++;
    int i = tp.second;
    pq.push({eval(i, groups[i]) - eval(i, groups[i] + 1), i});
    wanted--;
  }
  printf("%lld\n", outp);
  return 0;
}
