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;
template <typename T1, typename T2>
inline void chkmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chkmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
const int MAXN = 100228;
int n, k;
int a[MAXN];
int cur[MAXN];
ll get(ll x, ll cnt) {
  if (x <= cnt) {
    return x;
  }
  ll f = x / cnt;
  ll f1 = cnt - x % cnt;
  ll f2 = f + 1;
  ll f3 = x % cnt;
  return f * f * f1 + f2 * f2 * f3;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  set<pair<ll, int> > s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cur[i] = 1;
    s.insert(make_pair(get(a[i], cur[i]) - get(a[i], cur[i] + 1), i));
  }
  for (int i = 0; i < k - n; i++) {
    auto d = *s.rbegin();
    s.erase(d);
    cur[d.second]++;
    s.insert(make_pair(
        get(a[d.second], cur[d.second]) - get(a[d.second], cur[d.second] + 1),
        d.second));
  }
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    ans += get(a[i], cur[i]);
  }
  cout << ans << '\n';
}
