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;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  char c;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') {
    int x = (c = getchar()) - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return -x;
  }
  int x = c - '0';
  while ((c = getchar()) >= '0') x = x * 10 + c - '0';
  return x;
}
int64_t sqr(int64_t x) { return x * x; }
const int N = 1e5 + 5;
int a[N], c[N];
int64_t W(int a, int c) {
  int x = a / c;
  int b = a % c;
  return sqr(x) * (c - b) + sqr(x + 1) * b;
}
int main() {
  int n, k;
  cin >> n >> k;
  int64_t sum = 0;
  priority_queue<pair<int64_t, int>> heap;
  auto push = [&](int i) {
    if (c[i] < a[i]) heap.push({W(a[i], c[i]) - W(a[i], c[i] + 1), i});
  };
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    sum += sqr(a[i]);
    c[i] = 1;
    push(i);
  }
  for (int tmp = n + 1; tmp <= k; ++tmp) {
    auto pr = heap.top();
    heap.pop();
    sum -= pr.first;
    int i = pr.second;
    ++c[i];
    push(i);
  }
  cout << sum << endl;
}
