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 <cstring>
#include <vector>
#include <tuple>
#include <string>
#include <algorithm>

 
using namespace std;

typedef long long ll;
typedef long double ld;
#define L(ID) ((ID*2)+1)
#define R(ID) ((ID*2)+2)
#define M(L,R) ((L+R)/2)
#define forr(start,end) for(int i = start; i < end; i++)
#define gap ' '
 
const int MAXN = 100 + 5;
char s[MAXN];
ll n, t, k;

ll largest(ll a, ll b){
	return (a<b)?b:a;
}
ll smallest(ll a, ll b){
	if(a < b)return a;
	return b;
}


int main(){
	cin >> s >> k;
	bool isone = true;
	for(int i = 0; i < k; ++i){
		if(s[i] != '1'){
			isone = false;
			break;
		}
	}
	if(isone){
		cout << '1' << endl;
		return 0;
	}
	if(k == 1){
		cout << s[0] << endl;
		return 0;
	}
	for(int i = 0; i < strlen(s); ++i){
		if(s[i] != '1'){
			cout << s[i] << endl;
			return 0;
		}
	}
	cout << '1' << endl;

    return 0;
}