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 long long mod = 1e9 + 7;
int t, n, k;
long long a[1000010], b[1000010];
priority_queue<pair<long long, long long> > que;
long long cal(long long x, long long sz) {
  long long avg = x / sz;
  long long re = x % sz;
  return avg * avg * (sz - re) + (avg + 1) * (avg + 1) * re;
}
int main() {
  t = 1;
  while (t--) {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", &a[i]);
      b[i] = 1;
    }
    for (int i = 1; i <= n; ++i) {
      que.push(make_pair(cal(a[i], 1) - cal(a[i], 2), i));
    }
    for (int i = 1; i <= k - n; ++i) {
      pair<long long, long long> pa = que.top();
      que.pop();
      int idx = pa.second;
      b[idx]++;
      que.push(make_pair(cal(a[idx], b[idx]) - cal(a[idx], b[idx] + 1), idx));
    }
    long long ans = 0;
    while (!que.empty()) {
      pair<long long, long long> pa = que.top();
      que.pop();
      int idx = pa.second;
      ans += cal(a[idx], b[idx]);
    }
    printf("%lld\n", ans);
  }
}
