description: You are given an integer array a_1, a_2, ..., a_n and integer k.
  
  In one step you can 
  
    * either choose some index i and decrease a_i by one (make a_i = a_i - 1); 
    * or choose two indices i and j and set a_i equal to a_j (make a_i = a_j). 
  
  
  
  What is the minimum number of steps you need to make the sum of array ∑_{i=1}^{n}{a_i} ≤ k? (You are allowed to make values of array negative).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^{15}) — the size of array a and upper bound on its sum.
  
  The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array itself.
  
  It's guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k.
  
  Example
  
  Input
  
  
  4
  1 10
  20
  2 69
  6 9
  7 8
  1 2 1 3 1 2 1
  10 1
  1 2 3 1 2 6 1 6 8 10
  
  
  Output
  
  
  10
  0
  2
  7
  
  Note
  
  In the first test case, you should decrease a_1 10 times to get the sum lower or equal to k = 10.
  
  In the second test case, the sum of array a is already less or equal to 69, so you don't need to change it.
  
  In the third test case, you can, for example: 
  
    1. set a_4 = a_3 = 1; 
    2. decrease a_4 by one, and get a_4 = 0. 
  
  As a result, you'll get array [1, 2, 1, 0, 1, 2, 1] with sum less or equal to 8 in 1 + 1 = 2 steps.
  
  In the fourth test case, you can, for example: 
  
    1. choose a_7 and decrease in by one 3 times; you'll get a_7 = -2; 
    2. choose 4 elements a_6, a_8, a_9 and a_{10} and them equal to a_7 = -2. 
  
  As a result, you'll get array [1, 2, 3, 1, 2, -2, -2, -2, -2, -2] with sum less or equal to 1 in 3 + 4 = 7 steps.
solution: 
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.*;
  
  public class C {
      static class RealScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
      static long max = Long.MAX_VALUE;
  
      static long call(List<Long> list, long x, long sum, long k, long n) {
          long val = list.get(0) - x, tot = sum - x, ans = x;
          if (tot <= k) {
              return ans;
          }
          for (long i = n - 1; i > 0; i--) {
              tot += (val - list.get((int) i));
              ans++;
              if (k >= tot) {
                  return ans;
              }
          }
          return max;
      }
  
      public static void main(String[] args) {
  //        PrintWriter out = new PrintWriter(System.out);
          RealScanner sc = new RealScanner();
          int t = sc.nextInt();
          while (t-- > 0) {
              long n, k;
              n = sc.nextLong();
              k = sc.nextLong();
              long sum = 0;
  //            long[] a = new long[(int) n + 1];
              List<Long> list = new ArrayList<>();
              for (int i = 0; i < n; i++) {
                  long x = sc.nextLong();
                  list.add(x);
                  sum += x;
              }
              Collections.sort(list);
  //            long low = 0;           //first way
  //            long high = sum + k;
  //            long m1, m2;
  //            while (low < high) {
  //                long mid = (low + high) / 2;
  //                m1 = call(list, mid, sum, k, n);
  //                m2 = call(list, mid + 1, sum, k, n);
  //                if (m1 <= m2 && m1 != max) {
  //                    high = mid;
  //                } else {
  //                    low = mid + 1;
  //                }
  //            }
  //            System.out.println(call(list, low, sum, k, n));
  
              long tot = sum;            //2nd way
              ArrayDeque<Long> deque = new ArrayDeque<>();
              int i = 1;
              for (; i < n; i++) {
                  deque.addFirst(list.get(i) - list.get(0));
              }
              long count = 0;
              long min_val = Long.MAX_VALUE;
  
              while (tot > k && !deque.isEmpty()) {
                  long cal = (tot + count - k) / (count + 1);
                  min_val = Math.min(min_val, count + cal);
                  tot = tot - deque.getFirst();
                  count++;
                  deque.pop();
  //                System.out.println(tot + " " + s.pop()+" "+count);
              }
  //            System.out.println(tot);
  //            System.out.println(linkedList);
  //            System.out.println(min_val);
              if (tot > k) {
                  count += (tot - k + count) / (count + 1);
              }
              System.out.println(Math.min(count, min_val));
          }
  //        out.flush();
  //        out.close();
          System.out.close();
      }
  }
  