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 N = 100005;
int n, a[N], k;
long long cost(int len, int parts) {
  int len1 = len / parts;
  int len2 = len1 + 1;
  int cnt2 = len % parts, cnt1 = parts - cnt2;
  return (1LL * cnt1 * len1 * len1 + 1LL * cnt2 * len2 * len2);
}
struct node {
  int len, parts;
  node() { len = parts = 0; }
  node(int _len, int _parts) {
    len = _len;
    parts = _parts;
  }
  long long optimizerAnswer() const {
    return cost(len, parts) - cost(len, parts + 1);
  }
};
bool operator<(node a, node b) {
  return a.optimizerAnswer() < b.optimizerAnswer();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  priority_queue<node> pq;
  for (int(i) = (1); (i) <= (n); ++(i)) {
    cin >> a[i];
    pq.push(node(a[i], 1));
  }
  while (k-- > n) {
    node u = pq.top();
    pq.pop();
    pq.push(node(u.len, u.parts + 1));
  }
  long long ans = 0;
  while (pq.size()) {
    node u = pq.top();
    pq.pop();
    ans += cost(u.len, u.parts);
  }
  cout << ans;
}
