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>
int main(void) {
	long long a, b, c, k, h;
	int i, n, j;
	scanf("%lld", &a);
	scanf("%lld", &k);
	n = 0;
	b = a;
	for (i = 1; a >= 1; i++) {
		a = a / 10;
		n++;
	}
	
	long long  p[100];
	p[n] = b % 10;
	for (i = 1; i <= n - 1; i++) {
		c = 1;
		for (j = 1; j <= i; j++) {
			c = c * 10;
		}
		c = c * 10;
		p[n - i] = b % c - p[n - i + 1];
		c = c / 10;
		p[n - i] = p[n - i] / c;
	}
	h = 0;
	for (i = 1; p[i] == 1; i++) {
		h++;
	}
	if (h == 0) printf("%lld", p[1]);
	else if (k <= h) printf("%d", 1);
	else printf("%lld", p[h + 1]);
	return 0;

}