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;
long long MOD = 998244353;
const long long INF = 1e18;
const double E = 1e-18;
const double PI = acos(-1);
mt19937_64 rnd(time(0));
long long p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD,
          md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;
long long n, m, k, T, a[1000001], lst[100001], mx[1000001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x == 0 || y == 0) return x + y;
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
long long get(long long h, long long k) {
  long long o = h / k, p = h % k, y = k - p, u = o + 1;
  return 2 *
         (p * (p - 1) / 2 * u * u + y * (y - 1) / 2 * o * o + y * p * u * o);
}
void solve() {
  cin >> n >> k;
  set<pair<long long, int> > s;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx[i] = 1;
    if (a[i] > 1) s.insert(make_pair(-get(a[i], 2), i));
    ans += a[i] * a[i];
  }
  for (int i = 0; i < k - n; i++) {
    pair<long long, long long> x = (*s.begin());
    ans += x.first;
    s.erase(s.begin());
    mx[x.second]++;
    lst[x.second] -= x.first;
    if (mx[x.second] + 1 <= a[x.second]) {
      s.insert(make_pair(-get(a[x.second], mx[x.second] + 1) + lst[x.second],
                         x.second));
    }
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
