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;
int N, K;
long long a[100010];
long long calc(long long x, long long d) {
  long long dv = x / d, md = x % d;
  return (d - md) * dv * dv + md * (dv + 1) * (dv + 1);
}
struct cost {
  long long delta;
  long long v, d;
  friend bool operator<(cost c1, cost c2) { return c1.delta < c2.delta; }
  cost(long long _v, long long _d) {
    v = _v;
    d = _d;
    delta = calc(v, d) - calc(v, d + 1);
  }
};
priority_queue<cost> heap;
int main() {
  scanf("%d%d", &N, &K);
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &a[i]);
    ans += 1LL * a[i] * a[i];
  }
  for (int i = 1; i <= N; i++) {
    heap.push(cost(a[i], 1));
  }
  for (int fuck = 1; fuck <= K - N; fuck++) {
    cost tmp = heap.top();
    ans -= tmp.delta;
    heap.pop();
    heap.push(cost(tmp.v, tmp.d + 1));
  }
  printf("%lld\n", ans);
}
