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: # cook your dish here
  import sys
  #import random
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  from collections import deque
  #sys.setrecursionlimit(10**8)
  from queue import PriorityQueue as pq
  from math import gcd
  #import math
  input_ = lambda: sys.stdin.readline().strip("\r\n")
  ii = lambda : int(input_())
  il = lambda : list(map(int, input_().split()))
  ilf = lambda : list(map(float, input_().split()))
  lii = lambda : list(map(int, list(ip())))
  ip = lambda : input_()
  fi = lambda : float(input_())
  ap = lambda ab,bc,cd : ab[bc].append(cd)
  li = lambda : list(input_())
  pr = lambda x : print(x)
  prinT = lambda x : print(x)
  f = lambda : sys.stdout.flush()
  inv =lambda x:pow(x,mod-2,mod)
  dx = [0,0,1,-1]
  dy = [1,-1,0,0]
  mod = 10**9 + 7
  mod1 = 998244353
  
  
  for _ in range (ii()) :
      n,k = il()
      a = il()
      s = sum(a)
  
      if (s <= k) :
          print(0)
          continue
  
      if (n == 1) :
          print(s - k)
          continue
  
      a.sort()
      ans = 0
      ind = n - 1
  
      #print(s,a)
  
      while (ind > 0) :
          if (a[ind] == a[0]) :
              break
          s -= a[ind]
          s += a[0]
          ans += 1
          ind -= 1
  
          if (s <= k) :
              break
  
      if (s > k) :
          ans += s - k
  
      ans = min(ans,sum(a) - k)
  
      s = sum(a)
  
      t = 0
  
      for i in range (n-1,0,-1) :
          t1 = n - i
          t += a[i]
  
          c = (s - t + (a[0]*t1) - k + t1) // (t1 + 1)
  
          c = max(0,c)
          ans = min(c+t1,ans)
  
      print(ans)
  
      
      
          
          
  