Problem:
﻿Score: 300 points

Problem 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?

Constraints

S is a string of length between 1 and 100 (inclusive).

K is an integer between 1 and 10^{18} (inclusive).

The length of the string after 5 \times 10^{15} days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity's string after 5 \times 10^{15} days.

Sample Input 1

1214
4

Sample Output 1

2

The string S changes as follows:

Now: 1214

After one day: 12214444

After two days: 1222214444444444444444

After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5 \times 10^{15} days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

3
157

Sample Output 2

3

The initial string is 3. The string after 5 \times 10^{15} days consists only of 3.

Sample Input 3

299792458
9460730472580800

Sample Output 3

2

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def change(c)
  case c
  when '1'
    '1'
  when '2'
    '22'
  when '3'
    '333'
  when '4'
    '4444'
  when '5'
    '55555'
  when '6'
    '666666'
  when '7'
    '7777777'
  when '8'
    '88888888'
  when '9'
    '999999999'
  end
end

s = gets.chomp
k = gets.chomp.split(' ').map(&:to_i)[0]
if s.length == 1
  puts s[0]
else
  5000_000_000_000_000.times do
    ne = ''
    first = ''
    second = ''
    fin = true
    s.split('').each_with_index do |c, i|
      first = c if i.zero?
      second = c if first == '1' && second == '' && c != '1'
      fin = false if first == '1' && second != '' && second != c
      fin = false if first != '1' && first != c
      break if i > k
      ne += change(c)
    end
    s = ne
    break if fin 
  end
  puts s[k-1]
end