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 <class T>
inline T bigmod(T b, T p, T M) {
  T ret = 1;
  for (; p > 0; p >>= 1) {
    if (p & 1) ret = (ret * b) % M;
    b = (b * b) % M;
  }
  return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  unsigned long long int x = 1;
  while (p--) x *= b;
  return x;
}
const int mod = 1000000007;
long long int solve(long long int m, long long int Div) {
  long long int a = m / Div;
  long long int rem = m % Div;
  if (rem == 0) return Div * (a * a);
  long long int A = a, B = a + 1;
  return ((Div - rem) * A * A) + (rem * B * B);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long int n, k;
  cin >> n >> k;
  priority_queue<pair<long long int, pair<long long int, long long int> > > pq;
  long long int Tym = 0;
  for (int i = 1; i <= n; i++) {
    long long int L;
    cin >> L;
    Tym += L * L;
    long long int DecByOneCut = solve(L, 1) - solve(L, 2);
    pq.push({DecByOneCut, {L, 2}});
  }
  k = k - n;
  while (k--) {
    Tym -= pq.top().first;
    long long int L = pq.top().second.first, C = pq.top().second.second;
    pq.pop();
    long long int DecByOneCut = solve(L, C) - solve(L, C + 1);
    pq.push({DecByOneCut, {L, C + 1}});
  }
  cout << Tym << endl;
  return 0;
}
