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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
int main(void){
	char s[101];
	char k[101];
	char ans;
	scanf("%s",&s);
	scanf("%s",&k);
	if( strlen(k)>2){
		//kは100より大きい
		/*sを0から100まで見ていってずっと1だったら1を出力し、
		1以外の数値が来たらそれを出力する*/
		for(int i=0;i<100;i++){
			if( s[i] !='1'){
				printf("%c",s[i]);
				return 0;
			}
		}
		printf("1");
		return 0;
	}else{
		//kは100より小さい
		/*kを文字列から数字に直し、sを左からk番まで見ていって
		ずっと1だったら1を出力し、1以外の数値が来たらそれを出力する
		*/
		int x1=0;
		int x2=0;
		if (strlen(k)>1){
			x2 = k[1]-'0';
		}
		x1 = k[0]-'0';
		int x = x2*10 + x1;
		//printf("x = %d\n",x);
		for(int i=0;i<x;i++){
			if( s[i] !='1'){
				printf("%c",s[i]);
				return 0;
			}
		}
		printf("1");
		return 0;
	}
	return 0;
}