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;
inline int addmod(int a, int b) { return (a + b) % MOD; }
inline int mulmod(long long a, int b) { return (a * b) % MOD; }
inline int powmod(long long a, int x) {
  int ans = 1;
  while (x) {
    if (x & 1) {
      ans = (ans * a) % MOD;
    }
    x >>= 1;
    a = (a * a) % MOD;
  }
  return ans;
}
void init();
void solve();
int main() {
  init();
  solve();
  return 0;
}
const int MAXN = 1e5 + 5;
int n, k;
int a[MAXN];
struct Item {
  int val;
  int cnt;
  long long get_score() const {
    long long cur_ans = get_ans();
    Item tmp;
    tmp.val = val;
    tmp.cnt = cnt + 1;
    return cur_ans - tmp.get_ans();
  }
  long long get_ans() const {
    long long x = val / cnt;
    long long ans = (cnt - (val % cnt)) * x * x;
    ans += (val % cnt) * (x + 1) * (x + 1);
    return ans;
  }
  bool operator<(const Item &item) const {
    return get_score() < item.get_score();
  }
};
void init() {
  ios::sync_with_stdio(false);
  MOD = 1e9 + 7;
}
void solve() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  priority_queue<Item> que;
  for (int i = 0; i < n; i++) {
    Item item;
    item.val = a[i];
    item.cnt = 1;
    que.push(item);
  }
  for (int i = 0; i < k - n; i++) {
    Item item = que.top();
    que.pop();
    item.cnt++;
    que.push(item);
  }
  long long ans = 0;
  while (!que.empty()) {
    Item item = que.top();
    que.pop();
    ans += item.get_ans();
  }
  printf("%lld\n", ans);
}
