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>
const long long N = 1e5 + 51;
std::priority_queue<std::pair<long long, long long> > pq;
long long n, m, a[N], cnt[N];
inline long long f(long long x, long long c) {
  long long rem = x % c, ret = 0;
  ret += (x / c) * (x / c) * (c - rem);
  ret += (x / c + 1) * (x / c + 1) * rem;
  return ret;
}
long long ans;
signed main() {
  scanf("%lld%lld", &n, &m), m -= n;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    ans += a[i] * a[i];
    cnt[i] = 1;
    pq.push({f(a[i], 1) - f(a[i], 2), i});
  }
  while (m--) {
    auto [val, i] = pq.top();
    pq.pop();
    ans -= val, cnt[i]++;
    pq.push({f(a[i], cnt[i]) - f(a[i], cnt[i] + 1), i});
  }
  printf("%lld\n", ans);
}
