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 Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_macros)]
macro_rules ! read_value {($ next : expr , ($ ($ t : tt ) ,* ) ) => {($ (read_value ! ($ next , $ t ) ) ,* ) } ; ($ next : expr , [$ t : tt ; $ len : expr ] ) => {(0 ..$ len ) . map (| _ | read_value ! ($ next , $ t ) ) . collect ::< Vec < _ >> () } ; ($ next : expr , chars ) => {read_value ! ($ next , String ) . chars () . collect ::< Vec < char >> () } ; ($ next : expr , usize1 ) => {read_value ! ($ next , usize ) - 1 } ; ($ next : expr , $ t : ty ) => {$ next () . parse ::<$ t > () . expect ("Parse error" ) } ; }
#[allow(unused_macros)]
macro_rules ! input_inner {($ next : expr ) => {} ; ($ next : expr , ) => {} ; ($ next : expr , $ var : ident : $ t : tt $ ($ r : tt ) * ) => {let $ var = read_value ! ($ next , $ t ) ; input_inner ! {$ next $ ($ r ) * } } ; }
#[allow(unused_macros)]
macro_rules ! input {(source = $ s : expr , $ ($ r : tt ) * ) => {let mut iter = $ s . split_whitespace () ; let mut next = || {iter . next () . unwrap () } ; input_inner ! {next , $ ($ r ) * } } ; ($ ($ r : tt ) * ) => {let stdin = std :: io :: stdin () ; let mut bytes = std :: io :: Read :: bytes (std :: io :: BufReader :: new (stdin . lock () ) ) ; let mut next = move || -> String {bytes . by_ref () . map (| r | r . unwrap () as char ) . skip_while (| c | c . is_whitespace () ) . take_while (| c |! c . is_whitespace () ) . collect () } ; input_inner ! {next , $ ($ r ) * } } ; }

#[allow(non_snake_case)]
fn solve(S: &Vec<char>, K: u64) -> i32 {
    let last = K.min(S.len() as u64) as usize;
    for i in 0..last - 1 {
        if S[i] != '1' {
            return S[i].to_string().parse::<i32>().unwrap();
        }
    }
    if S[last - 1] != '1' {
        return S[last - 1].to_string().parse::<i32>().unwrap();
    }
    1
}

#[allow(non_snake_case)]
fn main() {
    input! {
        S: String,
        K: u64,
    }
    let ret = solve(&S.chars().collect(), K);
    println!("{}", ret);
}