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 gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
const int m4i[4] = {-1, 0, 1, 0}, m4j[4] = {0, 1, 0, -1};
const int m8i[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          m8j[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int mxn = 1e5 + 5;
int n, k;
long long cost, x;
priority_queue<pair<long long, pair<long long, long long> > > pq;
long long sq(long long chad) { return chad * chad; }
long long val(long long len, long long nums) {
  long long unit = len / nums;
  long long extra = len - unit * nums;
  return (nums - extra) * sq(unit) + extra * sq(unit + 1);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> x;
    cost += sq(x);
    pq.push(pair<long long, pair<long long, long long> >(
        val(x, 1) - val(x, 2), pair<long long, long long>(x, 2)));
  }
  for (int i = 0; i < k - n; i++) {
    cost -= pq.top().first;
    int a = pq.top().second.first, b = pq.top().second.second;
    pq.pop();
    pq.push(pair<long long, pair<long long, long long> >(
        val(a, b) - val(a, b + 1), pair<long long, long long>(a, b + 1)));
  }
  cout << cost << endl;
}
