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 N = 1e6 + 5;
int n, k, cur[N];
long long a[N], ans;
priority_queue<pair<long long, int> > q;
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
inline long long calc(int x, int t) {
  long long d = x / t, rest = x % t;
  return d * d * (t - rest) + (d + 1) * (d + 1) * rest;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    ans += (long long)a[i] * a[i];
    cur[i] = 1;
    q.push(make_pair((long long)a[i] * a[i] - calc(a[i], 2), i));
  }
  for (int i = 1; i <= k - n; ++i) {
    int now = q.top().second;
    ++cur[now];
    ans -= q.top().first;
    q.pop();
    q.push(make_pair(calc(a[now], cur[now]) - calc(a[now], cur[now] + 1), now));
  }
  cout << ans << "\n";
  return 0;
}
