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: Runtime Error)

//ダメコード
#include<stdio.h>
#include<math.h>
#include<stdlib.h>

typedef long long int ll;

ll powLL(ll a, ll b){
    return (ll)pow((double)a, (double)b);
}

int main(void){
    char s[101];
    scanf("%s", s);
    ll k;
    scanf("%lld", &k);

    int idx = 0;
    ll tmp = s[0] - '0';//こいつを出力すると、なぜかIllegal instructionになる
    ll cnt = 0;
    ll max = 5 * powLL(10, 15);
    ll maxK = powLL(10, 18);

    ll d[maxK][2];
    
    while(cnt < k){//ガバガバ
        if(tmp == 1){
            d[idx][0] = tmp;
            d[idx][1] = 1;
            idx++;
            tmp = s[idx] - '0';
            cnt++;
        }else{
            ll t = 1;
            while((powLL(tmp, t) < k) && (t <= max)){
                t++;
            }
            cnt += powLL(tmp, t);
            d[idx][0] = tmp;
            d[idx][1] = cnt;
            if(t != max+1){break;}
            idx++;
            tmp = s[idx] - '0';
        }
    }
    return 0;
}