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 sc = new Scanner(System.in);
		String a = sc.next();
		double b=sc.nextDouble();
		String end = "";
		double t = 5000 * 1000000 * 1000000;
		double cnt = 0;
		boolean flg = false;
		
//		System.out.println(t);
		
		String str[] = new String[100];
		
		for(int i=0;i<a.length();i++){
			str[i] = a.substring(i, i+1);
		}
		
		for(int i=0; i<100; i++){
			if (str[i] == null){
				break;
			}
			switch(str[i]){
			case "1":
				cnt++;
				if (cnt >b){
					System.out.println("1");
					flg = true;
				}
				break;
			case "2":
				for (double j=0;j<t;j++){
					cnt = cnt+2;
					if (cnt >b){
						System.out.println("2");
						flg = true;
						break;
					}
				}
				break;
			case "3":
				for (double j=0;j<t;j++){
					cnt = cnt+3;
					if (cnt >b){
						System.out.println("3");
						flg = true;
						break;
					}
				}
				break;
			case "4":
				for (double j=0;j<t;j++){
					cnt = cnt+4;
					if (cnt >b){
						System.out.println("4");
						flg = true;
						break;
					}
				}
				break;
			case "5":
				for (double j=0;j<t;j++){
					cnt = cnt+5;
					if (cnt >b){
						System.out.println("5");
						flg = true;
						break;
					}
				}
				break;
			case "6":
				for (double j=0;j<t;j++){
					cnt = cnt+6;
					if (cnt >b){
						System.out.println("6");
						flg = true;
						break;
					}
				}
				break;
			case "7":
				for (double j=0;j<t;j++){
					cnt = cnt+7;
					if (cnt >b){
						System.out.println("7");
						flg = true;
						break;
					}
				}
				break;
			case "8":
				for (double j=0;j<t;j++){
					cnt = cnt+8;
					if (cnt >b){
						System.out.println("8");
						flg = true;
						break;
					}
				}
				break;
			case "9":
				for (double j=0;j<t;j++){
					cnt = cnt+9;
					if (cnt >b){
						System.out.println("9");
						flg = true;
						break;
					}
				}
				break;
			default:
			}
			if (flg){
				break;
			}
		}
//		System.out.println("cnt="+cnt);
		
		
		
		
	}

}