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 long long inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
long long f(long long l, long long p) {
  long long w = l / p;
  long long p2 = l % p, p1 = p - p2;
  return p1 * w * w + p2 * (w + 1) * (w + 1);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  long long ans = 0;
  priority_queue<pair<long long, pair<long long, long long>>> pque;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ans += a[i] * a[i];
    pque.push({f(a[i], 1) - f(a[i], 2), {a[i], 2}});
  }
  for (long long i = 0; i < k - n; i++) {
    auto x = pque.top();
    pque.pop();
    ans -= x.first;
    long long l = x.second.first, p = x.second.second;
    pque.push({f(l, p) - f(l, p + 1), {l, p + 1}});
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
