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: from collections import deque, defaultdict, Counter
  
  import random
  import sys
  import os
  from io import BytesIO
  
  sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
  # sys.setrecursionlimit()
  
  MOD = 1000000007
  
  
  def calc(n, k, a):
      s = sum(a)
      if s <= k: return 0
      if n == 1: return max(0, a[0] - k)
  
      a.sort()
      m = a[0]
      ans = max(0, s - k)
      cnt = 0
      
  
      for i in range(len(a)-1, 0, -1):
          s += m - a[i]
          cnt += 1
          # s <= k with cnt of m's
          tmp = (s - k + cnt) // (cnt + 1)
          #print(s - k + cnt)
          #print(i, f"{s = :}", ans, "=> ", cnt, tmp)
          ans = min(ans, cnt + max(0, tmp))
      
      return ans
  
  def solve():
      global tcases
      #n = int(sys.stdin.readline())
      n, k = map(int, sys.stdin.readline().split())
      a = list(map(int, sys.stdin.readline().split()))
      # s = sys.stdin.readline()[:-1]
      # p = sys.stdin.readline()[:-1]
      
      print(calc(n, k, a))
  
  def test():
      for _z in range(100000):
          n, k = [random.randint(5, 20) for _ in range(2)]
          a = [random.randint(1, 10) for _ in range(n)]
          calc(n, k, a)
  
  tcases = int(sys.stdin.readline())
  
  if tcases < 0:
      test()
      exit(0)
  
  for t in range(tcases):
      solve()
  
  
  