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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int cost(long long int val, long long int par) {
  long long int k = val / par;
  if (val % par) {
    return (par - val % par) * (k) * (k) + (k + 1) * (k + 1) * (val % par);
  } else {
    return k * k * par;
  }
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> a[i];
  set<pair<pair<long long int, long long int>, long long int>> s;
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += a[i] * a[i];
    s.insert({{-cost(a[i], 1) + cost(a[i], 2), 2}, i});
  }
  long long int rem = k - n;
  while (rem--) {
    auto temp = *s.begin();
    ans += temp.first.first;
    s.erase(s.begin());
    long long int i = temp.second;
    long long int p = temp.first.second;
    s.insert({{-cost(a[i], p) + cost(a[i], p + 1), p + 1}, i});
  }
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test;
  test = 1;
  while (test--) {
    solve();
    cout << '\n';
  }
  cerr << "Time : "
       << 1000 * ((long double)clock()) / (long double)CLOCKS_PER_SEC << "ms\n";
  return 0;
}
