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 <stdlib.h>

void getInput(char *pstr, int size)
{
    fgets(pstr, size, stdin);
}

int main() {
    char str[100];
    char *pstr = &str[0];

    char S[100];
    unsigned long K = 0;
    int result = 0;
    
    /* Sの読み込み */
    getInput(S, sizeof(S));
    //printf("%s\n", S);
    
    /* Kの読み込み */
    getInput(pstr, sizeof(str));
    K = atoi(pstr);
    //printf("%lu\n", K);
    
    /* 10^18 < 2^60なので左から1以外を探す */
    int count = 0;
    while (1) {
        result = S[count] - 48;
        if (1 != result) {
            break;
        }
        
        if (K == (count + 1)) {
            result = 1;
            break;
        }
        count++;
    }
    printf("%d\n", result);
    
    return 0;
}