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: Accepted)

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		String S = scan.next();
		long K = scan.nextLong();
		scan.close();
		int l = S.length();
		int []num = new int[S.length()];
		for(int i = 0; i < l; i++) {
			num[i] = Integer.parseInt(S.substring(i, i + 1));
		}
//		long[]k = new long[S.length()];
		long sum = l;
		if(sum >= K) {
			int cnt1 = 0;
			for(int i = 0; i < l; i++) {
				if(num[i] == 1) {
					cnt1++;
				}else {
					break;
				}
			}
			if(cnt1 >= K) {
				System.out.println(1);
				System.exit(0);
			}
		}
		long cnt = 1;

		int ans = 0;
		for(int i = 0; i < l; i++) {
			if(num[i] != 1) {
				ans = num[i];
				break;
			}
		}
		System.out.println(ans);

	}
}