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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;

using System.Collections;

class Program
{
    static void Main()
    {
        string S = Console.ReadLine();

        int K = int.Parse(Console.ReadLine());

        //2は、5000兆日後には2の5000兆乗の長さになる。
        //つまり、2以上も、5000兆日後にはnの5000兆乗の長さになる。

        //Kはせいぜい1以上10^18以下



        //文字列中の、最初の1が連続する部分の長さ
        var xxx = S.TakeWhile(itm => itm == '1').Count();

        //1が連続する部分の次の文字
        var yyy = S.SkipWhile(itm => itm == '1').Take(1);

        if (K <= xxx)
        {
            Console.WriteLine(1);
        }
        else
        {
            Console.WriteLine(yyy.Single());
        }
    }
}



static class MathUtil
{
    public static IEnumerable<IEnumerable<int>> GetPermutation(int n)
    {
        if (n == 0)
        {
            yield return Single(0);
            yield break;
        }

        for (int i = 0; i <= n; i++)
        {
            foreach (var x in GetPermutation(n - 1))
            {
                yield return Single(i).Concat(x.Select(itm => itm < i ? itm : itm + 1));
            }
        }
    }

    private static IEnumerable<int> Single(int x)
    {
        yield return x;
    }

    public static IEnumerable<int> EnumeratePrimeNumbers()
    {
        List<int> primes = new List<int>() { 2 };
        yield return 2;

        for (int i = 3; i < int.MaxValue; i++)
        {
            bool isPrime = true;
            int Sqrt = (int)Math.Sqrt(i) + 1;

            foreach (var prime in primes)
            {
                if (prime > Sqrt)
                {
                    break;
                }
                if (i % prime == 0)
                {
                    isPrime = false;
                    break;
                }
            }

            if (isPrime)
            {
                primes.Add(i);
                yield return i;
            }
        }
    }
}