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 int INF = 0x3f3f3f3f;
template <class T>
void read(T &val) {
  T x = 0;
  T bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int mod = 998244353;
const int maxn = 1e6 + 10;
int n, m, a[maxn], k, t;
char s[maxn];
priority_queue<pair<long long, pair<int, int>>> q;
long long cal(long long len, long long num) {
  long long unit = len / num;
  long long ext = len - unit * num;
  long long cost = unit * unit * (num - ext) + (unit + 1) * (unit + 1) * ext;
  return cost;
}
int main() {
  read(n);
  read(k);
  for (int i = 1; i <= n; i++) {
    long long x;
    read(x);
    q.push({cal(x, 1) - cal(x, 2), {x, 2}});
  }
  k -= n;
  while (k--) {
    auto u = q.top();
    q.pop();
    int tim = u.second.second, num = u.second.first;
    q.push({cal(num, tim) - cal(num, tim + 1), {num, tim + 1}});
  }
  long long ans = 0;
  while (!q.empty()) {
    auto u = q.top();
    q.pop();
    ans += cal(u.second.first, u.second.second - 1);
  }
  printf("%lld\n", ans);
  return 0;
}
