output_description: Print the number of times Takahashi will hear his favorite sound.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A B C
attempt: import sys
  
  sys.setrecursionlimit(10**8)
  
  
  def ii():
      return int(sys.stdin.readline())
  
  
  def mi():
      return map(int, sys.stdin.readline().split())
  
  
  def li():
      return list(map(int, sys.stdin.readline().split()))
  
  
  def li2(N):
      return [list(map(int, sys.stdin.readline().split())) for _ in range(N)]
  
  
  def dp2(ini, i, j):
      return [[ini] * i for _ in range(j)]
  
  
  def dp3(ini, i, j, k):
      return [[[ini] * i for _ in range(j)] for _ in range(k)]
  
  
  # import bisect #bisect.bisect_left(B, a)
  # from collections import defaultdict #d = defaultdict(int) d[key] += value
  # from itertools import accumulate #list(accumulate(A))
  
  A, B, C = mi()
  
  if A * C < B:
      print(C)
  else:
      print(B // A)
  
problem_description: Statement
  
  Takahashi likes the sound when he buys a drink from a vending machine.
  
  That sound can be heard by spending A yen (the currency of Japan) each time.
  
  Takahashi has B yen. He will hear the sound as many times as he can with that
  money, but at most C times, as he would be satisfied at that time.
  
  How many times will he hear the sound?
samples: [{"input": "2 11 4", "output": "4\n    \n\nSince he has not less than 8 yen, he will hear the sound four times and be\nsatisfied.\n\n* * *"}, {"input": "3 9 5", "output": "3\n    \n\nHe may not be able to be satisfied.\n\n* * *"}, {"input": "100 1 10", "output": "0"}]