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 N = 100030;
long long vt[N];
int cnt[N];
long long get(long long v, long long cnt) {
  long long v1 = v / cnt, v2 = v1 + 1;
  long long cnt2 = v % cnt, cnt1 = cnt - cnt2;
  return v1 * v1 * cnt1 + v2 * v2 * cnt2;
}
void solve() {
  int n, k;
  cin >> n >> k;
  set<pair<long long, int> > st;
  long long ans = 0;
  for (int id = 0; id < n; id++) {
    cin >> vt[id];
    ans += vt[id] * vt[id];
    cnt[id] = 1;
    if (vt[id] != 1) {
      st.insert({-get(vt[id], cnt[id]) + get(vt[id], cnt[id] + 1), id});
    }
  }
  k -= n;
  while (k--) {
    int id = st.begin()->second;
    ans += st.begin()->first;
    st.erase(*st.begin());
    cnt[id]++;
    if (vt[id] != cnt[id]) {
      st.insert({-get(vt[id], cnt[id]) + get(vt[id], cnt[id] + 1), id});
    }
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}
