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;
int n, k;
long long a[500005];
priority_queue<pair<pair<long long, long long>, int> > q;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  k -= n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    long long ans = a[i] * a[i] - (a[i] / 2) * (a[i] / 2) -
                    (a[i] / 2 + a[i] % 2) * (a[i] / 2 + a[i] % 2);
    q.push(pair<pair<long long, long long>, int>(
        pair<long long, long long>(ans, 1ll), i));
  }
  long long rlt = 0;
  while (k--) {
    pair<pair<long long, long long>, int> pp = q.top();
    pair<long long, long long> p = pp.first;
    q.pop();
    long long nn = p.second + 1;
    int id = pp.second;
    long long ll = a[id] / nn;
    long long r = a[id] % nn;
    long long rlt = 0;
    rlt += 1ll * r * (ll + 1) * (ll + 1);
    rlt += 1ll * (nn - r) * ll * ll;
    nn++;
    ll = a[id] / nn;
    r = a[id] % nn;
    rlt -= 1ll * r * (ll + 1) * (ll + 1);
    rlt -= 1ll * (nn - r) * ll * ll;
    q.push(pair<pair<long long, long long>, int>(
        pair<long long, long long>(rlt, nn - 1), id));
  }
  long long cur = 0;
  while (!q.empty()) {
    pair<pair<long long, long long>, int> pp = q.top();
    pair<long long, long long> p = pp.first;
    q.pop();
    long long nn = p.second;
    int id = pp.second;
    long long ll = a[id] / nn;
    long long r = a[id] % nn;
    cur += 1ll * r * (ll + 1) * (ll + 1);
    cur += 1ll * (nn - r) * ll * ll;
  }
  cout << cur << endl;
}
