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 mod = 1e9 + 7;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 10;
int d[N], a[N];
priority_queue<pair<long long, int> > q;
long long x2(long long x) { return x * x; }
long long cal(long long x, long long k) {
  long long val = x / k;
  long long num1 = x % k;
  long long num0 = k - num1;
  return num0 * x2(val) + num1 * x2(val + 1);
}
long long solve() {
  pair<long long, int> x = q.top();
  q.pop();
  int id = x.second;
  if (x.first == 1) return 0;
  long long tmp = x.first;
  q.push({cal(a[id], d[id]) - cal(a[id], ++d[id]), id});
  return x.first;
}
int main() {
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; i++) cin >> a[i], ans += 1ll * a[i] * a[i], d[i] = 2;
  for (int i = 1; i <= n; i++) {
    q.push({1ll * x2(a[i]) - x2(a[i] / 2) - x2(a[i] - a[i] / 2), i});
  }
  k -= n;
  while (k--) {
    ans -= solve();
  }
  cout << ans;
}
