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: Wrong Answer)

use proconio::{input, fastout};
use proconio::marker::{Chars};
use std::cmp::*;

macro_rules! debug {
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
  }

#[fastout]
fn main() {
    input!{
        s: Chars,
        k: usize,
    }

    let mut ans = 0;
    for i in 0..s.len() {
        if s[i]=='1' {
            continue;
        } else {
            ans = s[i].to_digit(10).unwrap();
            break;
        }
    }
    if ans==0 {
        ans = 1;
    }

    println!("{}", ans);

}