output_description: Print the K-th character from the left in Mr. Infinity's string after 5 \times
  10^{15} days.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      S
      K
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  from collections import defaultdict
  from collections import deque
  from functools import lru_cache
  
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i]) * n**i for i in range(len(str(X))))
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  S = S()
  K = I()
  n = len(S)
  ind = 0
  for i in range(n):
      if S[i] != "1":
          ind = i
          num = S[i]
          break
  if K > ind:
      P(num)
  else:
      P(1)
  
problem_description: Statement
  
  Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time
  the date changes, this string changes as follows:
  
    * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`.
  
  For example, if S is `1324`, it becomes `1333224444` the next day, and it
  becomes `133333333322224444444444444444` the day after next. You are
  interested in what the string looks like after 5 \times 10^{15} days. What is
  the K-th character from the left in the string after 5 \times 10^{15} days?
samples: [{"input": "1214\n    4", "output": "2\n    \n\nThe string S changes as follows:\n\n  * Now: `1214`\n  * After one day: `12214444`\n  * After two days: `1222214444444444444444`\n  * After three days: `12222222214444444444444444444444444444444444444444444444444444444444444444`\n\nThe first five characters in the string after 5 \\times 10^{15} days is\n`12222`. As K=4, we should print the fourth character, `2`.\n\n* * *"}, {"input": "3\n    157", "output": "3\n    \n\nThe initial string is `3`. The string after 5 \\times 10^{15} days consists\nonly of `3`.\n\n* * *"}, {"input": "299792458\n    9460730472580800", "output": "2"}]