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 MOD = 1e9 + 7;
long long int INF = 1e18;
long long int dx[] = {1, -1, 0, 0};
long long int dy[] = {0, 0, 1, -1};
long long int fun(long long int l, long long int n) {
  long long int a = l / n;
  long long int b = l % n;
  return (n - b) * a * a + b * (a + 1) * (a + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  priority_queue<pair<long long int, pair<long long int, long long int>>> st;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (a[i] * a[i]);
    st.push({fun(a[i], 1) - fun(a[i], 2), {a[i], 2}});
  }
  k -= n;
  while (k--) {
    ans -= st.top().first;
    long long int b = st.top().second.first, c = st.top().second.second;
    st.pop();
    st.push({fun(b, c) - fun(b, c + 1), {b, c + 1}});
  }
  cout << ans << endl;
  return 0;
}
