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;
mt19937 rnd(time(nullptr));
const long long INF = 1e18, P = 1e9 + 7, N = 2e5 + 10;
const long double EPS = 1e-9;
long long f(long long l, long long k) {
  if (k == 1) {
    return l * l;
  }
  long long x = l / k, ost = l % k;
  return (k - ost) * x * x + ost * (x + 1) * (x + 1);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  k -= n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  set<pair<long long, long long> > st;
  vector<long long> cur(n, 1);
  for (long long i = 0; i < n; i++) {
    st.insert({f(a[i], cur[i]) - f(a[i], cur[i] + 1), i});
  }
  for (long long w = 0; w < k; w++) {
    pair<long long, long long> p = *--st.end();
    st.erase(p);
    long long i = p.second;
    cur[i]++;
    st.insert({f(a[i], cur[i]) - f(a[i], cur[i] + 1), i});
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += f(a[i], cur[i]);
  }
  cout << ans << "\n";
  return 0;
}
