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 (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::*;

pub fn read_from<T: FromStr, R: Read>(s: &mut R) -> T {
    let s = s.bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    if !s.is_empty() {
        if let Ok(s) = s.parse::<T>() {
            s
        } else {
            panic!("faild to parse {}", s);
        }
    } else {
        if cfg!(local) {
            // eprintln!("reached to EOF");
        }
        std::process::exit(0)
    }
}

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let mut stdin = stdin.lock();
    read_from(&mut stdin)
}

pub fn readn<T: FromStr>(n: usize) -> Vec<T> {
    let stdin = stdin();
    let mut stdin = stdin.lock();
    (0..n).map(|_| read_from(&mut stdin)).collect()
}

const N: usize = 5000_0000_0000_0000;

fn calc(c: char) -> usize {
    if c == '1' {
        1
    } else {
        N + N
    }
}

fn main() {
    let s: Vec<char> = read::<String>().chars().collect();
    let n = s.len();
    let k: usize = read();
    let k = k - 1;
    let mut len = 0;
    let mut ans = '?';
    for i in 0..n {
        len += calc(s[i]);
        ans = s[i];
        if len >= k {
            break;
        }
    }
    println!("{}", ans);
}