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 maxn = 1e5 + 5, inf = 0x3f3f3f3f, mod = 1e9 + 7;
const double eps = 1e-6;
long long ans = 0;
int n, k;
int a[maxn];
long long cal(long long x, long long num) {
  long long a = x % num, b = num - a;
  return a * (x / num + 1) * (x / num + 1) + b * (x / num) * (x / num);
}
struct node {
  long long x, num;
  friend bool operator<(node a, node b) {
    long long cost1 = cal(a.x, a.num) - cal(a.x, a.num + 1);
    long long cost2 = cal(b.x, b.num) - cal(b.x, b.num + 1);
    return cost1 < cost2;
  }
};
priority_queue<node> pq;
signed main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pq.push({a[i], 1});
    ans += 1ll * a[i] * a[i];
  }
  int now = n;
  while (now < k) {
    node temp = pq.top();
    pq.pop();
    now++;
    ans -= cal(temp.x, temp.num);
    temp.num++;
    ans += cal(temp.x, temp.num);
    pq.push(temp);
  }
  printf("%lld\n", ans);
  return 0;
}
