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 cnt = 0;
vector<int> cand;
bool prime[100000];
void SieveOfEratosthenes(int n) {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
int prob = 0;
void dfs(vector<int> adj[], int v, bool visited[], stack<int>& s,
         bool visited2[]) {
  visited[v] = true;
  int i;
  int c = 1;
  bool f = true;
  for (i = 0; i < adj[v].size(); i++) {
    if (!visited[adj[v][i]]) {
      f = false;
      dfs(adj, adj[v][i], visited, s, visited2);
    }
  }
  if (adj[v].size() == 1) {
    s.push(v);
    visited2[v] = true;
  }
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long add(long long x, long long y) {
  x += y;
  while (x >= 1000000007) x -= 1000000007;
  while (x < 0) x += 1000000007;
  return x;
}
long long mul(long long x, long long y) { return (x * 1ll * y) % 1000000007; }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long inv(long long x) { return binpow(x, 1000000007 - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
long long fact[1000001];
void precalc() {
  fact[0] = 1;
  for (long long i = 1; i < 1000001; i++) fact[i] = mul(fact[i - 1], i);
}
long long C(long long n, long long k) {
  if (n == 0) return 1;
  return divide(fact[n], mul(fact[k], fact[n - k]));
}
int dfs(bool visited[], vector<int> adj[], int v) {
  int c = 1;
  visited[v] = true;
  for (int i = 0; i < adj[v].size(); i++) {
    if (!visited[adj[v][i]]) {
      c += dfs(visited, adj, adj[v][i]);
    }
  }
  return c;
}
struct comp {
  bool operator()(const pair<int, int>& p1, const pair<int, int>& p2) {
    if (p1.first == p2.first) {
      return p1.second < p2.second;
    } else
      return p1.first > p2.first;
  }
};
long long sub(long long A, long long B) {
  return (A - B + 1000000007) % 1000000007;
}
void toposort(vector<int>& order, int v, int visited[], vector<int> adj[]) {
  visited[v] = 1;
  int i;
  for (i = 0; i < adj[v].size(); i++) {
    if (!visited[adj[v][i]]) {
      toposort(order, adj[v][i], visited, adj);
    }
  }
  order.push_back(v);
}
long long val(long long len, long long nums) {
  long long unit = len / nums;
  long long extra = len - unit * nums;
  return (nums - extra) * (unit) * (unit) + extra * (unit + 1) * (unit + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(5);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    multiset<long long> s;
    priority_queue<pair<long long, pair<long long, long long> > > pq;
    long long cost = 0;
    for (int i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      cost += x * x;
      pq.push({val(x, 1) - val(x, 2), {x, 2}});
    }
    int c = 1;
    while (c <= k - n) {
      c++;
      cost -= pq.top().first;
      long long a = pq.top().second.first;
      long long b = pq.top().second.second;
      pq.pop();
      pq.push({val(a, b) - val(a, b + 1), {a, b + 1}});
    }
    cout << cost << "\n";
  }
  return 0;
}
