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;
using ll = long long;
ll n, k;
vector<ll> as;
vector<ll> cur_parts;
priority_queue<pair<ll, ll>> pq;
inline ll compute_cost(ll len, ll parts) {
  ll n_longer = len % parts;
  ll n_shorter = parts - n_longer;
  ll longer = (len / parts + 1);
  ll shorter = (len / parts);
  return n_shorter * shorter * shorter + n_longer * longer * longer;
}
inline ll compute_gain(ll len, ll parts) {
  return compute_cost(len, parts) - compute_cost(len, parts + 1);
}
void input() {
  cin >> n >> k;
  for (ll i = 0; i < n; i++) {
    ll a;
    cin >> a;
    as.push_back(a);
  }
  cur_parts.resize(n, 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  input();
  for (int i = 0; i < n; i++) {
    if (cur_parts[i] < as[i]) {
      pq.push({compute_gain(as[i], cur_parts[i]), i});
    }
  }
  for (int i = n; i < k; i++) {
    ll cut = pq.top().second;
    pq.pop();
    cur_parts[cut]++;
    if (cur_parts[cut] < as[cut]) {
      pq.push({compute_gain(as[cut], cur_parts[cut]), cut});
    }
  }
  ll res = 0;
  for (int i = 0; i < n; i++) {
    res += compute_cost(as[i], cur_parts[i]);
  }
  cout << res << "\n";
  return 0;
}
