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>
template <class T>
bool chmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
using namespace std;
const int INF = INT_MAX;
const long long LNF = LONG_LONG_MAX;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
long long sq(long long i) { return i * i; }
long long cost(int len, int cut) {
  long long seg = len / cut;
  long long rem = len % cut;
  return sq(seg) * (cut - rem) + rem * sq(seg + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n], b[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long second = k - n;
    priority_queue<pair<long long, long long>> p;
    long long c = 0;
    for (long long i = 0; i < n; i++) c += a[i] * a[i];
    for (long long i = 0; i < n; i++) b[i] = 1;
    for (long long i = 0; i < n; i++) {
      long long d = cost(a[i], 2) - cost(a[i], 1);
      p.push(make_pair(-d, i));
    }
    for (long long i = 0; i < second; i++) {
      pair<long long, long long> x = p.top();
      p.pop();
      c -= x.first;
      ++b[x.second];
      long long d =
          cost(a[x.second], b[x.second] + 1) - cost(a[x.second], b[x.second]);
      p.push(make_pair(-d, x.second));
    }
    cout << c << '\n';
  }
  return 0;
}
