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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define faster ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
 
typedef vector<int> vi;;
typedef vector<ll> vll;

#define l() length()
#define sz() size()
 
#define b() begin()
#define e() end()
#define all(x) x.begin(), x.end()
 
#define pb(i) push_back(i)
#define PB() pop_back()
 
 
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;
#define mp make_pair
#define Fi first
#define Se second
 
const int maxx=1e6+9;
 
// Happy Coding!

int _find(string str){
    
    for(int i=0;i<str.l();i++){
        if(str[i]!='1'){return i;}
    }
    return -1;
}

void solve(){
    string str;
    ll k;
    cin>>str;
    cin>>k;
    
    int p=_find(str);
    if(p==-1){
        cout<<1<<endl;
        return ;
    }
    else{
        k--;
        if(k>p){
            cout<<str[p];
        }
        else{
            cout<<1;
        }
    }
}

int main() {
    faster;
    int t=1;
    // cin>>t;
    while(t-->0){
        solve();
    }
	return 0;
}