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 + 10;
const int Inf = 0x7f7f7f7f;
const long long Inf_ll = 1ll * Inf * Inf;
const int Mod = 1e9 + 7;
const double eps = 1e-7;
long long ans = 0;
int n, k, a[Maxn];
priority_queue<pair<long long, pair<int, int>>> q;
pair<pair<int, int>, pair<int, int>> Div(int x, int m) {
  if (m >= x) return make_pair(make_pair(1, x), make_pair(0, 0));
  pair<int, int> a, b;
  a.first = x / m;
  b.first = x / m + 1;
  b.second = x - m * a.first;
  a.second = (x - b.first * b.second) / a.first;
  return make_pair(a, b);
}
long long calc(pair<pair<int, int>, pair<int, int>> pp) {
  pair<int, int> p1 = pp.first, p2 = pp.second;
  return 1ll * p1.first * p1.first * p1.second +
         1ll * p2.first * p2.first * p2.second;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ans += 1ll * a[i] * a[i];
    pair<pair<int, int>, pair<int, int>> pp = Div(a[i], 2);
    long long tmp = calc(pp);
    q.push(make_pair(1ll * a[i] * a[i] - tmp, make_pair(a[i], 1)));
  }
  for (int i = n + 1; i <= k; i++) {
    if (q.empty()) break;
    pair<long long, pair<int, int>> plp = q.top();
    q.pop();
    ans -= plp.first;
    pair<int, int> p = plp.second;
    long long Now = calc(Div(p.first, p.second + 1)),
              New = calc(Div(p.first, p.second + 2));
    q.push(make_pair(Now - New, make_pair(p.first, p.second + 1)));
  }
  printf("%lld\n", ans);
  return 0;
}
