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;
const int INF = 1e9 + 100;
ll f(ll a, ll b) {
  ll low = a / b;
  ll lowCnt = b - a % b;
  ll highCnt = a % b;
  return low * low * lowCnt + (low + 1) * (low + 1) * highCnt;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  set<pair<ll, ll>> vals;
  vector<ll> a(n);
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ans += a[i] * a[i];
    assert(f(a[i], 1) == a[i] * a[i]);
    vals.emplace(f(a[i], 1) - f(a[i], 2), i);
  }
  vector<ll> splitCnt(n, 1);
  for (int i = 0; i < k - n; i++) {
    if (vals.empty()) {
      cout << "foo;\n";
      return 0;
    }
    auto it = *vals.rbegin();
    vals.erase(it);
    ans -= it.first;
    int id = it.second;
    splitCnt[id]++;
    vals.emplace(f(a[id], splitCnt[id]) - f(a[id], splitCnt[id] + 1), id);
  }
  cout << ans << '\n';
  return 0;
}
