description: Monocarp is playing yet another computer game. In this game, his character has to kill a dragon. The battle with the dragon lasts 100^{500} seconds, during which Monocarp attacks the dragon with a poisoned dagger. The i-th attack is performed at the beginning of the a_i-th second from the battle start. The dagger itself does not deal damage, but it applies a poison effect on the dragon, which deals 1 damage during each of the next k seconds (starting with the same second when the dragon was stabbed by the dagger). However, if the dragon has already been poisoned, then the dagger updates the poison effect (i.e. cancels the current poison effect and applies a new one).
  
  For example, suppose k = 4, and Monocarp stabs the dragon during the seconds 2, 4 and 10. Then the poison effect is applied at the start of the 2-nd second and deals 1 damage during the 2-nd and 3-rd seconds; then, at the beginning of the 4-th second, the poison effect is reapplied, so it deals exactly 1 damage during the seconds 4, 5, 6 and 7; then, during the 10-th second, the poison effect is applied again, and it deals 1 damage during the seconds 10, 11, 12 and 13. In total, the dragon receives 10 damage.
  
  Monocarp knows that the dragon has h hit points, and if he deals at least h damage to the dragon during the battle — he slays the dragon. Monocarp has not decided on the strength of the poison he will use during the battle, so he wants to find the minimum possible value of k (the number of seconds the poison effect lasts) that is enough to deal at least h damage to the dragon.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
  
  The first line of the test case contains two integers n and h (1 ≤ n ≤ 100; 1 ≤ h ≤ 10^{18}) — the number of Monocarp's attacks and the amount of damage that needs to be dealt.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9; a_i < a_{i + 1}), where a_i is the second when the i-th attack is performed.
  
  Output
  
  For each test case, print a single integer — the minimum value of the parameter k, such that Monocarp will cause at least h damage to the dragon.
  
  Example
  
  Input
  
  
  4
  2 5
  1 5
  3 10
  2 4 10
  5 3
  1 2 4 5 7
  4 1000
  3 25 64 1337
  
  
  Output
  
  
  3
  4
  1
  470
  
  Note
  
  In the first example, for k=3, damage is dealt in seconds [1, 2, 3, 5, 6, 7].
  
  In the second example, for k=4, damage is dealt in seconds [2, 3, 4, 5, 6, 7, 10, 11, 12, 13].
  
  In the third example, for k=1, damage is dealt in seconds [1, 2, 4, 5, 7].
solution: # import collections
  # import random
  # import math
  # import itertools
  # import math
  # mport math
  # import bisect
  from collections import defaultdict
  # import itertools
  # from sys import stdin, stdout
  # import math
  # import math
  # import bisect
  # import bisect
  # import collections
  # import itertools
  # import math
  import sys
  
  # import operator
  # from decimal import Decimal
  
  # sys.setrecursionlimit(10**6)
  
  class DisjointSet:
      parent = {}
      cp = {}
      # stores the depth of trees
      rank = {}
      m = 0
  
      # perform MakeSet operation
      def makeSet(self, universe):
          # create `n` disjoint sets (one for each item)
          for i in universe:
              self.parent[i] = i
              self.cp[i] = 1
              self.rank[i] = 0
  
      # Find the root of the set in which element `k` belongs
      def Find(self, k):
          # if `k` is not the root
          if self.parent[k] != k:
              # path compression
              self.parent[k] = self.Find(self.parent[k])
          return self.parent[k]
  
      # Perform Union of two subsets
      def Union(self, a, b):
          # find the root of the sets in which elements `x` and `y` belongs
          x = self.Find(a)
          y = self.Find(b)
  
          # if `x` and `y` are present in the same set
          if x == y:
  
              return
  
          # Always attach a smaller depth tree under the root of the deeper tree.
          if self.rank[x] > self.rank[y]:
              self.parent[y] = x
              self.cp[x] += self.cp[y]
              self.cp[y] = 0
              self.m = max(self.m, self.cp[x])
          elif self.rank[x] < self.rank[y]:
              self.parent[x] = y
              self.cp[y] += self.cp[x]
              self.cp[x] = 0
              self.m = max(self.m, self.cp[y])
          else:
              self.parent[x] = y
              self.rank[y] = self.rank[y] + 1
              self.cp[y] += self.cp[x]
              self.cp[x] = 0
              self.m = max(self.m, self.cp[y])
  
  
  def printSets(universe, ds):
      print([ds.Find(i) for i in universe])
  
  
  p2D = lambda x: print(*x, sep="\n")
  
  
  def II(): return int(sys.stdin.buffer.readline())
  
  
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  
  
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  
  
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  
  
  def BI(): return sys.stdin.buffer.readline().rstrip()
  
  
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  
  
  def li(): return [int(i) for i in input().split()]
  
  
  def lli(rows): return [li() for _ in range(rows)]
  
  
  def si(): return input()
  
  
  def ii(): return int(input())
  
  
  def ins(): return input().split()
  
  
  # mod = 10**9+7
  # fval = [1]
  # for i in range(1,200001):
  #     x = (fval[-1]*i)%mod
  #     fval.append(x)
  #
  # ifac = [1]*(200001)
  # ifac[200000] = pow(fval[200000], -1, mod)
  # for i in range(200000-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%mod
  
  
  def drank(d, processing, da, rank):
      tmp = 10 ** 9
      if len(d[da]) == 1:
          return 1
      for di in d[da]:
          if processing[di - 1] == 0:
              processing[di - 1] = 1
              tmp = min(tmp, drank(d, processing, di, rank))
              processing[di - 1] = 0
      rank[da - 1] = tmp + 1
      return tmp + 1
  
  
  
  def binary_search(n, a):
      l = len(a)
      low = 0
      high = l-1
      while high>=low:
          mid = (high+low)//2
          if a[mid] == n:
              return mid
          elif a[mid]>n:
              high = mid-1
          else:
              low = mid+1
      return -1
  
  is_prime = [True]*(10**6+1)
  def primeseive(n):
      is_prime[0] = is_prime[1] = False
      i = 2
      while i*i<=n:
          if is_prime[i] == True:
              for j in range(i*i,n+1,i):
                  is_prime[j] = False
          i+=1
  
  def satis(a, m):
      i = 0
      l = len(a)
      j = 1
      ans = 0
      while i<l:
          k = 0
          while j<l and a[j]< a[i]+m:
              j+=1
              k+=1
          if k==0:
              ans += m
              i+=1
              j+=1
          else:
              ans += (a[i+k]- a[i])
              i = i+k
              j = i+1
      return ans
  
  def solve():
      n , h = li()
      a = li()
      if n == 1:
          return h
      if h<=n:
          return 1
      else:
          low = 1
          high = h
          m = 10**18
  
          while high>=low:
              mid = (low+high)//2
              x = satis(a,mid)
              if x>h:
                  high = mid-1
                  if x<m:
                      m = x
                      prev = mid
              elif x<h:
                  low = mid+1
              else:
                  return mid
          return prev
  
  
  
  
  
  
  def main():
      #primeseive(10**6)
      #print(is_prime)
      for _ in range(ii()):
      #solve()
          sys.stdout.write(str(solve()) + "\n")
      # print(solve())
  
      #     z += str(ans) + '\n'
      # print(len(ans), ' '.join(map(str, ans)), sep='\n')
      # stdout.write(z)
  
  
  # for interactive problems
  # print("? {} {}".format(l,m), flush=True)
  # or print this after each print statement
  # sys.stdout.flush()
  
  
  if __name__ == "__main__":
      main()