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;
long long a[100010];
long long s[100010];
long long split(long long first, long long n) {
  long long q = first / n;
  long long r = first % n;
  return r * (q + 1) * (q + 1) + (n - r) * q * q;
}
long long get_bon(long long first, long long cnt) {
  if (cnt == first) return -1e15;
  return split(first, cnt) - split(first, cnt + 1);
}
long long get_loss(long long first, long long cnt) {
  if (cnt == 1) return 1e15;
  return split(first, cnt - 1) - split(first, cnt);
}
int main() {
  long long n, k;
  cin >> n >> k;
  int need_bon = k - n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    s[i] = 1;
    while (need_bon > 0 && s[i] < a[i]) {
      s[i]++;
      need_bon--;
    }
  }
  priority_queue<pair<long long, pair<long long, long long>>,
                 vector<pair<long long, pair<long long, long long>>>,
                 less<pair<long long, pair<long long, long long>>>>
      plus;
  priority_queue<pair<long long, pair<long long, long long>>,
                 vector<pair<long long, pair<long long, long long>>>,
                 greater<pair<long long, pair<long long, long long>>>>
      minus;
  for (int i = 0; i < n; i++) {
    minus.push({get_loss(a[i], s[i]), {i, s[i]}});
    plus.push({get_bon(a[i], s[i]), {i, s[i]}});
  }
  while (!plus.empty() && !minus.empty()) {
    while (!plus.empty() &&
           (plus.top().second.second != s[plus.top().second.first]))
      plus.pop();
    while (!minus.empty() &&
           (minus.top().second.second != s[minus.top().second.first]))
      minus.pop();
    if (minus.empty() || plus.empty()) break;
    while (!minus.empty() &&
           plus.top().second.first == minus.top().second.first)
      minus.pop();
    while (!minus.empty() &&
           (minus.top().second.second != s[minus.top().second.first]))
      minus.pop();
    if (minus.empty()) break;
    if (plus.top().first <= minus.top().first) break;
    int pl = plus.top().second.first;
    int mn = minus.top().second.first;
    s[pl]++;
    s[mn]--;
    minus.push({get_loss(a[pl], s[pl]), {pl, s[pl]}});
    plus.push({get_bon(a[pl], s[pl]), {pl, s[pl]}});
    minus.push({get_loss(a[mn], s[mn]), {mn, s[mn]}});
    plus.push({get_bon(a[mn], s[mn]), {mn, s[mn]}});
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += split(a[i], s[i]);
  }
  cout << ans << endl;
}
