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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function Main(input) {
    input = input.trim().split("\n").map(function(x) { return x.split(" ")});
    let S = input[0][0];
    let K = input[1][0];
    K = BigInt(K);
    let ans;
    if (S[0] !== '1') ans = S[0];
    else {
        let flag;
        let tmpLength = 0n;
        ans = 1;
        for (let i = 0; i < S.length; i++){
            flag = false;
            if (S[i] !== '1') {
                ans = S[i];
                break;
            }
            tmpLength = tmpLength + 1n;
            flag = true;
        }

        if (!flag){
            if (K <= tmpLength) ans = 1;
        }
    }

    console.log(ans);

}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));