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 n;
pair<long long, int> t[400007];
void build(long long a[], int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = {a[tl], tl};
  } else {
    int tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
    if (t[v * 2].first > t[v * 2 + 1].first) {
      t[v] = t[v * 2];
    } else {
      t[v] = t[2 * v + 1];
    }
  }
}
pair<long long, int> maxx(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    return {0, -1};
  }
  if (l == tl && r == tr) {
    return t[v];
  }
  int tm = (tl + tr) / 2;
  pair<int, int> l1 = maxx(v * 2, tl, tm, l, min(r, tm)),
                 r1 = maxx(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  if (l1.first > r1.first) {
    return l1;
  }
  return r1;
}
void update(int v, int tl, int tr, int pos, long long new_val) {
  if (tl == tr) {
    t[v] = {new_val, pos};
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm) {
      update(v * 2, tl, tm, pos, new_val);
    } else {
      update(v * 2 + 1, tm + 1, tr, pos, new_val);
    }
    if (t[v * 2].first > t[v * 2 + 1].first) {
      t[v] = t[2 * v];
    } else {
      t[v] = t[2 * v + 1];
    }
  }
}
long long pro1(pair<int, int> f) {
  long long p = f.first / f.second;
  long long k = p * f.second;
  long long h = f.first - p * f.second;
  return (h * (p + 1) * (p + 1) + p * p * (f.second - h));
}
long long pro(pair<int, int> f) {
  long long p = f.first / (f.second + 1);
  long long k = p * (f.second + 1);
  long long h = f.first - p * (f.second + 1);
  return pro1(f) - (h * (p + 1) * (p + 1) + p * p * (f.second + 1 - h));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > q(n);
  for (int i = 0; i < n; ++i) {
    cin >> q[i].first;
    q[i].second = 1;
  }
  long long a[n];
  for (int i = 0; i < n; i++) {
    a[i] = pro(q[i]);
  }
  build(a, 1, 0, n - 1);
  for (int i = 0; i < k - n; i++) {
    pair<long long, int> g = maxx(1, 0, n - 1, 0, n - 1);
    int pos = g.second;
    q[pos].second++;
    a[pos] = pro(q[pos]);
    update(1, 0, n - 1, pos, a[pos]);
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    res += pro1(q[i]);
  }
  cout << res << "\n";
}
