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 <string>
#include <vector>
#include <list>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <cctype>
#include <fstream>

using namespace std;

#define ll long long
#define INF 2000000000
#define LINF 9000000000000000000
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define du(x) cout<<#x<<'='<<x<<endl
#define du2(x,y) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<'\n'<<endl
#define du3(x,y,z) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<' '<<#z<<'='<<z<<'\n'<<endl
#define darr(arr,n) rep(i,(n-arr)){cout<<*(arr+i)<<' ';}cout<<endl
#define ddarr(arr,a,b) rep(i,a){rep(j,b){cout<<*(arr+i*b+j)<<' ';}cout<<endl;}
#define ddd ifstream in("Text.txt");cin.rdbuf(in.rdbuf();

int main() {
    string S;
    ll K;
    cin >> S >> K;
    int i = 0;
    while (i < K) {
        if (S[i] != '1') {
            cout << S[i] << endl;
            return 0;
        }
        i++;
    }
    cout << 1 << endl;
}