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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long a,b=0,c,d,e[1000001],f[1000001];
string a1;
inline long long read(){
	char s;
	long long x,f=1,o=0;
	while(s<'0'||s>'9'){
		if(s=='-'){
			f=-1;
		}
		s=getchar();
	}
	while(s>='0'&&s<='9'){
		o=o*10+s-'0';
		s=getchar();
	}
	return x=o*f;	
}
int main(){
	cin>>a1;
	a=read();
	for(long long i=0;i<a1.length();i++){
		if(a1[i]=='1'){
			b+=1;
			if(b>=a){
				printf("%lld\n",a1[i]-'0');
				return 0;
			}			
		}		
		else{
			c=a1[i]-'0';
			d=1;
			for(long long j=0;j<5000000000000000;j++){
				d*=c;
				b+=d;
				if(b>=a){
					printf("%lld\n",a1[i]-'0');
					return 0;
				}				
			}
		}
	}
	return 0;
}