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")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long MOD = 1e+9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
const long long N = 3e+5 + 7;
vector<long long> adj[N];
long long vis[N] = {};
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
int t = 1;
long long n, k, a, ans;
long long split(long long x, long long n) {
  long long len = x / n;
  return (x % n) * (len + 1) * (len + 1) + (n - x % n) * len * len;
}
struct compare {
  bool operator()(const pair<long long, long long> &i,
                  const pair<long long, long long> &j) const {
    return split(i.first, i.second) - split(i.first, i.second + 1) >
           split(j.first, j.second) - split(j.first, j.second + 1);
  }
};
void MAIN() {
  cin >> n >> k;
  multiset<pair<long long, long long>, compare> set1;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    set1.insert({x, 1});
  }
  for (long long i = 1; i <= k - n; i++) {
    auto it = *set1.begin();
    set1.erase(set1.begin());
    it.second++;
    set1.insert(it);
  }
  for (auto it : set1) {
    long long x = it.first;
    long long n = it.second;
    ans += split(x, n);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  ;
  while (t--) {
    MAIN();
  }
}
