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;
using vi = vector<int>;
using pii = pair<int, int>;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int n, k;
ll a[maxn];
ll cal(ll x, ll num) {
  ll a1 = x / num, n2 = x % num;
  ll n1 = num - n2, a2 = a1 + 1;
  return a1 * a1 * n1 + a2 * a2 * n2;
}
priority_queue<pair<ll, pair<ll, ll>>> q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  ll sum = 0;
  for (int i = (1); i <= (int)(n); i++) {
    cin >> a[i];
    sum += a[i] * a[i];
  }
  for (int i = (1); i <= (int)(n); i++)
    q.push({cal(a[i], 1) - cal(a[i], 2), {a[i], 1}});
  int m = k - n;
  while (m--) {
    auto now = q.top();
    q.pop();
    sum -= now.first;
    ll sum = now.second.first, cnt = now.second.second;
    q.push({cal(sum, cnt + 1) - cal(sum, cnt + 2), {sum, cnt + 1}});
  }
  cout << sum << '\n';
  return 0;
}
