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()
{
	int S[100], K;
	int i;
	int A;

	scanf("%d", &S);
	for (i = 0;i++;i < 100)
	{
		if (S[i] == 1)
		{
			S[i] = 1;
		}
		if (S[i] == 2)
		{
			S[i] = S[i] * 10 + 2;
		}
		if (S[i] == 3)
		{
			S[i] = S[i] * 100 + 33;
		}
		if (S[i] == 4)
		{
			S[i] = S[i] * 1000 + 444;
		}
		if (S[i] == 5)
		{
			S[i] = S[i] * 10000 + 5555;
		}
		if (S[i] == 6)
		{
			S[i] = S[i] * 100000 + 66666;
		}
		if (S[i] == 7)
		{
			S[i] = S[i] * 1000000 + 777777;
		}
		if (S[i] == 8)
		{
			S[i] = S[i] * 10000000 + 8888888;
		}
		if (S[i] == 9)
		{
			S[i] = S[i] * 100000000 + 99999999;
		}
	}

	scanf("%d", &K);
	if (K > 0 && K < 1000000000000000000)
	{
		A = S[K];
		printf("%d", A);
	}
	else
	{
		printf("error\n");
	}

	return 0;
}