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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <cmath>
#include <stdio.h>
#include <list>
#include <numeric>
#include <stack>
#include <queue>
#include <tuple>
#include <bitset>
#include <map>
#include <math.h>
//　ceil(a/b)   (a + (b - 1))/ b
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); i++)
typedef long long ll;
typedef pair<int,int> P;
const int inf=1000000007;
const int mod=1000000007;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};

int gcd(int a, int b){
  if (a%b==0){
    return(b);
  }
  else{
    return(gcd(b,a%b));
  }
}


// ll memo[2020][2020];
// int pascal_triangle(){
//   for(int i = 0; i < 2020; i++){
//     for(int j = 0; j < i+1; j++){
//       if(j == 0 || j == i) memo[i][j] = 1;
//       else{
// 	memo[i][j] = (memo[i-1][j-1]+memo[i-1][j]);
//       }
//     }
//   }
//   return 0;
// }


// int factrial(int n){
//   if(n == 0) return 1;
//   return n*factrial(n-1);
// }

// int divisor(int a){
//   int cnt = 0;
//   for(int i=1; i <= a; i++) if(a%i == 0) cnt++;
//   return cnt;
// }

// int comb(int n, int k){
//   return fact(n) / (fact(n-k)*fact(k));
// }



int main(){
  string s;
  ll K;
  ll num1 = 0;
  cin >> s >> K;
  rep(i,s.size()) if(s[i] == '1') num1++;
  if(K != 1 && s[0] == '1' && K <= num1) cout << s[1] << endl;
  else if(K > num1) cout << s[num1] << endl;
  else cout << s[0] << endl;
}