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;
int mod = 1000000007;
const int inf = 1034567891;
const long long LL_INF = 1234567890123456789ll;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T GCD(T a, T b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
template <typename T>
string toString(T a) {
  return to_string(a);
}
template <typename T>
void toInt(string s, T& x) {
  stringstream str(s);
  str >> x;
}
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
long long get(long long v, long long c) {
  long long d = v / c;
  long long r = v % c;
  long long a = c - r;
  long long b = r;
  long long ans = (d * d) * a + ((d + 1) * (d + 1)) * b;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  long long v;
  multiset<pair<long long, pair<int, int>>,
           greater<pair<long long, pair<int, int>>>>
      s;
  for (int i = 0; i < n; i++) {
    cin >> v;
    s.insert({get(v, 1) - get(v, 2), {1, v}});
  }
  int need = k - n;
  for (int i = 0; i < need; i++) {
    auto it = *s.begin();
    s.erase(s.begin());
    long long v = it.second.second, c = it.second.first + 1;
    long long prev = get(v, c), cur = get(v, c + 1);
    s.insert({prev - cur, {c, v}});
  }
  long long ans = 0;
  for (auto it : s) {
    ans += get(it.second.second, it.second.first);
  }
  cout << ans << '\n';
  return 0;
}
