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 sys
  read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
  #import bisect,string,math,time,functools,random,fractions
  #from bisect import*
  #from heapq import heappush,heappop,heapify
  #from collections import deque,defaultdict,Counter
  #from itertools import permutations,combinations,groupby
  rep=range;R=range
  def I():return int(input())
  def LI():return [int(i) for i in input().split()]
  def SLI():return sorted([int(i) for i in input().split()])
  def LI_():return [int(i)-1 for i in input().split()]
  def S_():return input()
  def IS():return input().split()
  def LS():return [i for i in input().split()]
  def NI(n):return [int(input()) for i in range(n)]
  def NI_(n):return [int(input())-1 for i in range(n)]
  def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
  def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
  def StoLI():return [ord(i)-97 for i in input()]
  def ItoS(n):return chr(n+97)
  def LtoS(ls):return ''.join([chr(i+97) for i in ls])
  def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
  def RI(a=1,b=10):return random.randint(a,b)
  def INP():
      N=1000
      n=random.randint(2,N)
      mn=0
      mx=30
      #a=[random.randint(mn,mx) for i in range(n)]
      return str(n)
  def Rtest(T):
      case,err=0,0
      for i in range(T):
          inp=INP()
          a1=naive(inp)
          a2=solve(inp)
          if a1!=a2:
              print(inp)
              print('naive',a1)
              print('solve',a2)
              err+=1
          case+=1
      print('Tested',case,'case with',err,'errors')
  def GI(V,E,ls=None,Directed=False,index=1):
      org_inp=[];g=[[] for i in range(V)]
      FromStdin=True if ls==None else False
      for i in range(E):
          if FromStdin:
              inp=LI()
              org_inp.append(inp)
          else:
              inp=ls[i]
          if len(inp)==2:a,b=inp;c=1
          else:a,b,c=inp
          if index==1:a-=1;b-=1
          aa=a;bb=b;g[a].append(bb)
          if not Directed:g[b].append(aa)
      return g,org_inp
  def RE(E):
      rt=[[]for i in range(len(E))]
      for i in range(len(E)):
          for nb,d in E[i]:
              rt[nb]+=(i,d),
      return rt
  def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
      #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
      mp=[boundary]*(w+2);found={}
      for i in R(h):
          s=input()
          for char in search:
              if char in s:
                  found[char]=((i+1)*(w+2)+s.index(char)+1)
                  mp_def[char]=mp_def[replacement_of_found]
          mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
      mp+=[boundary]*(w+2)
      return h+2,w+2,mp,found
  def TI(n):return GI(n,n-1)
  def accum(ls):
      rt=[0]
      for i in ls:rt+=[rt[-1]+i]
      return rt
  def bit_combination(n,base=2):
      rt=[]
      for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
      return rt
  def gcd(x,y):
      if y==0:return x
      if x%y==0:return y
      while x%y!=0:x,y=y,x%y
      return y
  def YN(x):print(['NO','YES'][x])
  def Yn(x):print(['No','Yes'][x])
  def show(*inp,end='\n'):
      if show_flg:print(*inp,end=end)
  
  mo=10**9+7
  #mo=998244353
  inf=float('inf')
  FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
  alp=[chr(ord('a')+i)for i in range(26)]
  #sys.setrecursionlimit(10**7)
  
  
  show_flg=False
  show_flg=True
  
  ans=0
  
  
  
  for _ in range(I()):
      ans=0
      n,h=LI()
      a=LI()
      
      def c(k):
          d=0
          for i in range(n):
              if i<n-1:
                  d+=min(a[i+1]-a[i],k)
              else:
                  d+=k
          #show(k,d)
          return d>=h
      
      l,r=0,h+1
      while r-l>1:
          m=l+r>>1
          if c(m):
              r=m
          else:
              l=m
      
      print(r)
  
  
  