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;
inline long long read() {
  long long x = 0;
  bool f = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x * 10) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct abc {
  long long v, k, s;
  bool operator<(const abc &p) const {
    if (v == p.v) return (s % k) < p.s % p.k;
    return v < p.v;
  }
};
long long f(long long x, long long y) {
  if (y == 1) return x * x;
  long long c = x % y, z = x / y;
  return c * (z + 1) * (z + 1) + (y - c) * z * z;
}
priority_queue<abc> q;
long long n, k, ans;
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; i++) {
    long long x = read();
    q.push((abc){f(x, 1) - f(x, 1 + 1), 1, x});
  }
  for (long long i = 1; i <= k - n; i++) {
    abc s = q.top();
    q.pop();
    s.k++;
    s.v = f(s.s, s.k) - f(s.s, s.k + 1);
    q.push(s);
  }
  while (!q.empty()) {
    abc s = q.top();
    q.pop();
    long long c = s.s % s.k;
    for (long long i = 1; i <= c; i++) {
      ans += (s.s / s.k + 1) * (s.s / s.k + 1);
    }
    for (long long i = c + 1; i <= s.k; i++) {
      ans += (s.s / s.k) * (s.s / s.k);
    }
  }
  cout << ans;
}
