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;
long long int mod = 998244353;
long long int INF = 1e9;
long long int po(long long int a, long long int b) {
  long long int re = 1;
  while (b) {
    if (b % 2 == 1) {
      re = (re * a);
      re %= mod;
    }
    a = (a * a);
    a %= mod;
    b >>= 1;
  }
  return re;
}
long long int vl(long long int a, long long int b) {
  long long int c = a / b;
  return (a % b) * (c + 1) * (c + 1) + (b - a % b) * c * c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  while (test--) {
    int n, k;
    cin >> n >> k;
    long long int a[n], ans = 0;
    set<pair<long long int, int> > c;
    pair<long long int, int> tem;
    int cn[n];
    for (long long int i = (0); i < (n); i++) {
      cin >> a[i];
      cn[i] = 1;
      ans += a[i] * a[i];
      c.insert(make_pair(vl(a[i], 1) - vl(a[i], 2), i));
    }
    for (long long int j = (n); j < (k); j++) {
      tem = *(--c.end());
      c.erase(tem);
      ans -= tem.first;
      int i = tem.second;
      cn[i]++;
      c.insert(make_pair(vl(a[i], cn[i]) - vl(a[i], cn[i] + 1), i));
    }
    cout << ans;
  }
  return 0;
}
