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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		char[] s = in.nextLine().toCharArray();
		long k = Long.parseLong(in.nextLine());
		System.out.println(solve(s, k));
		in.close();
	}

	private static char solve(char[] s, long k) {
		int index = getIndex(s);
		if (index < 0) {
			// すべて1で構成されている
			return '1';
		} else if (index > k) {
			// １の範囲
			return '1';
		} else {
			return s[index];
		}
	}

	private static int getIndex(char[] s) {
		int result = -1;
		for (int i = 0; i < s.length; ++i) {
			if (s[i] != '1') {
				result = i;
				break;
			}
		}
		return result;
	}
}