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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val s = readLine()!!
    val k = readLine()!!.toLong()

    // 5000兆 -> 5 x 10^15

    // 先頭に１があってそれを使う場合 or 長さが１の場合
    if (s.length == 1) {
        println(s[0])
        return
    }

    // WA 同じ数字が何回でも出て良い -> １が連続するケース を考慮漏れ
//    if (s.first() == '1') {
//
//        if (k == 1L) {
//            println(s.first())
//            return
//        }
//
//        println(s[1])
//        return
//    }
//
//    println(s[0])

    // ケース分類
    // 12 : 1が最初だけ -> 12222222222222222・・・
    // 11112 : 1が複数最初からつながる -> 1111222222222222222222222・・・
    // 23 : １以外が最初にある -> 22222222222222222・・・
    // 2 : 一文字 -> 22222222222222・・・

    // -> 最初の状態で k or 最後の文字目までに 1 以外を見つけたらそれを出力，なければ１を出力

    for (i in 0 until k) {
        if (s[i.toInt()] != '1') {
            println(s[i.toInt()])
            return
        }

        if(i.toInt() == (s.length-1)){
            break
        }
    }

    println(1)
}