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 DIM = 1e5 + 5;
struct usu {
  long long ch;
  long long bef;
  int k, i;
  bool operator<(const usu &aux) const {
    if (ch != aux.ch) return ch > aux.ch;
    return i < aux.i;
  }
};
int n, k;
int a[DIM];
priority_queue<usu> pq;
void solve() {
  long long sol = 0;
  for (int i = 1; i <= n; ++i) {
    int x = a[i] / 2, y = a[i] / 2 + a[i] % 2;
    sol = sol + 1LL * a[i] * a[i];
    pq.push({1LL * x * x + 1LL * y * y - 1LL * a[i] * a[i],
             1LL * x * x + 1LL * y * y, 2, i});
  }
  int cnt = k - n;
  while (cnt > 0) {
    --cnt;
    long long nv = pq.top().ch, bef = pq.top().bef;
    int k = pq.top().k, i = pq.top().i;
    pq.pop();
    ++k;
    int x = a[i] / k, y = a[i] / k + 1;
    int rest = a[i] % k;
    sol = sol + nv;
    long long aux = 1LL * x * x * (k - rest) + 1LL * y * y * rest;
    pq.push({aux - bef, aux, k, i});
  }
  printf("%lld", sol);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  solve();
  return 0;
}
