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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

import sys
import itertools

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline

in_n = lambda: int(readline())
in_nn = lambda: map(int, readline().split())
in_nl = lambda: list(map(int, readline().split()))
in_na = lambda: map(int, read().split())
in_s = lambda: readline().rstrip().decode('utf-8')


def main():

    S = list(map(int, in_s()))
    K = in_n()

    if not(set(S) - {1}):
        print(1)
        exit()

    gr = itertools.groupby(S)

    init_s = []
    for k, l in gr:
        init_s.append((k, len(list(l))))

    N = len(init_s)

    for t in range(100):
        for i in range(N):
            k, v = init_s[i]
            if k == 1:
                continue
            else:
                init_s[i] = (k, k * v)

    tv = 0
    for k, v in init_s:
        tv += v
        if tv > K:
            ans = k
            break

    print(ans)


if __name__ == '__main__':
    main()