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: Wrong Answer)

#include<stdio.h>

//#define N 5000000000000000
#define N 50

long int pow(int base,int exp);

int main(void)
{
	char S[100];
	int i,j,moji=0;
	long int K,sum=0,count[100][10];
	
	scanf("%s",S);
	scanf("%ld",&K);
	
	for(i=1;i<=9;i++){
		for(j=1;j<=moji;j++){
			count[j][i]=0;
		}
	}
	
	for(i=0;S[i]!='\0';i++){
		switch(S[i]){
			case '1':
				count[i+1][1]++;
				break;
			case '2':
				count[i+1][2]++;
				break;
			case '3':
				count[i+1][3]++;
				break;
			case '4':
				count[i+1][4]++;
				break;
			case '5':
				count[i+1][5]++;
				break;
			case '6':
				count[i+1][6]++;
				break;
			case '7':
				count[i+1][7]++;
				break;
			case '8':
				count[i+1][8]++;
				break;
			case '9':
				count[i+1][9]++;
				break;
		}
		moji++;
	}
	
	for(i=1;i<=9;i++){
		for(j=1;j<=moji;j++){
			count[j][i]*=pow(i,10);
		}
	}
	//printf("%ld\n",pow(3,5));
	
	for(i=1;i<=9;i++){
		for(j=1;j<=moji;j++){
			sum+=count[j][i];
			if(sum>=K){
				printf("%d\n",i);
				return 0;
			}
		}
	}
	
	return 0;
}

long int pow(int base,int exp)
{
	/*if(exp==0){
		return 1;
	}
	return base*pow(base,exp-1);*/
	int i,sum=1;
	
	for(i=1;i<=exp;i++){
		sum*=base;
	}
	return sum;
}