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 Counter
  from sys import *
  from collections import defaultdict as dd
  from math import *
  
  # sys.stdin = open('input.txt', 'r')  
  # sys.stdout = open('output.txt', 'w')
  
  #input funcoions
  def inp():
      return int(stdin.readline().strip())
  def vinp():
      return map(int,stdin.readline().strip().split())
  def linp():
      return list(map(int,stdin.readline().strip().split()))
  def sinp(n = 1):
      if n==1:
          return stdin.readline().strip()
      elif n==2:
          return list(stdin.readline().strip())
      else:
          return list(stdin.readline().split())
  
  #output funcoion
  def pr(*x, end = "\n"):
      print(*x,end = end)
  
  #others
  def mod(f, val = 1000000007):
      return f % val
  def csort(c):
      sorted(c.items(), key=lambda pair: pair[1], reverse=True)
  def indc(l,n):
      c={}
      for i in range(n):
          c[l[i]]=c.get(l[i],[])+[i+1]
      return c
  
  if __name__ =="__main__":
      cou=inp()
      for _ in range(cou):
          n ,k= vinp()
          l = linp()
          s = sum(l)
          l.sort()
          l1 = [l[i]-l[0] for i in range(1,n)]
          co = 0
          mi = maxsize
          while l1!=[] and s>k:
              mi = min(mi , co + (s - k + co )//(co + 1))
              s -= l1[-1]
              l1.pop(-1)
              co+=1
          if s>k:
              co += (s - k + co)//(co + 1);
          pr(min(mi,co))