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>
#pragma GCC optimize("Ofast,unroll-loops,fast-math,inline,no-stack-protector")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,avx,avx2,abm,mmx,popcnt")
using namespace std;
long long cut(long long x, long long n) {
  long long lowest = x / n;
  long long highest = lowest + 1;
  long long cnthighest = x % n;
  long long cntlowest = n - cnthighest;
  return lowest * lowest * cntlowest + highest * highest * cnthighest;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n == k) {
    long long ans = 0;
    for (long long &el : a) {
      ans += el * el;
      ;
    }
    cout << ans << endl;
    return 0;
  }
  sort(a.begin(), a.end());
  vector<long long> cnt(n, 1);
  set<pair<long long, long long>> ss;
  for (long long i = 0; i < n; i++) {
    if (a[i] != 1) {
      ss.insert({cut(a[i], 2) - cut(a[i], 1), i});
    }
  }
  long long need = k - n;
  while (need) {
    need--;
    auto pp = *ss.begin();
    ss.erase(ss.begin());
    cnt[pp.second]++;
    if (cnt[pp.second] != a[pp.second]) {
      ss.insert({cut(a[pp.second], cnt[pp.second] + 1) -
                     cut(a[pp.second], cnt[pp.second]),
                 pp.second});
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += cut(a[i], cnt[i]);
  }
  cout << ans << endl;
}
