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 sys
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
  p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  # dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
  # dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
  # inf = 18446744073709551615
  inf = 4294967295
  # md = 10**9+7
  md = 998244353
  
  def solve():
      n, k = LI()
      aa = LI()
      if sum(aa) <= k:
          print(0)
          return
      aa.sort()
      s = 0
      ans = inf
      for i in range(1, n):
          d = k-s
          c = n+1-i
          x = d//c
          cur = max(0, aa[0]-x)+c-1
          if cur < ans: ans = cur
          s += aa[i]
      x = k-s
      cur = max(0, aa[0]-x)
      if cur < ans: ans = cur
      print(ans)
  
  for testcase in range(II()):
      solve()
  