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, collections, math, bisect, heapq, random, functools,io,os,copy
  input = sys.stdin.readline
  out = sys.stdout.flush
  def rep():
      a = list(map(int,input().split()))
      return a
  
  def sep():
      a = list(input().split())
      return a
  
  
  class UnionFind:
      def __init__(self, x) -> None:
          self.uf = [-1] * x
  
      def find(self, x):
          r = x
          while self.uf[x] >= 0:
              x = self.uf[x]
  
          while r != x:
              self.uf[r], r = x, self.uf[r]
          return x
  
      def union(self, x, y):
          ux, uy = self.find(x), self.find(y)
          if ux == uy:
              return
          if self.uf[ux] >= self.uf[uy]:
              self.uf[uy] += self.uf[ux]
              self.uf[ux] = uy
          else:
              self.uf[ux] += self.uf[uy]
              self.uf[uy] = ux
          return
  
      def __print__(self):
          return self.uf
  
  
  def spfa(x,g,n):
      dis = [float('inf') for i in range(n)]
      dis[x] = 0
      state = [False for i in range(n)]
      state[x] = True
      queue = collections.deque()
      queue.append(x)
      while queue:
          cur = queue.popleft()
          state[cur] = False
          for next_ in g[cur]:
              if dis[next_] > dis[cur] + 1:
                  dis[next_] = dis[cur] + 1
                  if state[next_] == False:
                      state[next_] = True
                      queue.append(next_)
      return dis
  
  
  def gcd(x,y):
      if y == 0:
          return x
      return gcd(y,x % y)
  
  
  def valid(x,a,v,total,k):
      # 枚举减少几次
      minv = v[0]
      n = len(v)
      if x > n - 1:
         minv = minv - (x - n + 1)
         total = total - (x - n + 1)
         x = n - 1
  
      for i in range(x + 1):
          sufsum = a[x - i]
          cursum = total - i - sufsum + (x - i) * (minv - i)
          if cursum <= k:
              return True
  
      return False
  def solve():
      n,k = map(int,input().split())
      a = rep()
      t = sum(a)
      a.sort()
      suf = [0]
      for i in range(n - 1,-1,-1):
          suf.append(suf[-1] + a[i])
  
      if t <= k:
          print(0)
      else:
  
          l,r = 1,t - k
          while l <= r:
              mid = l + (r - l) // 2
              if valid(mid,suf,a,t,k):
                  r = mid - 1
              else:
                  l = mid + 1
          print(r + 1)
  
  
  
  if __name__ == '__main__':
      for i in range(int(input())):
          solve()