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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

/*
  sen 1,000
  man 1,000,000
  oku 1,000,000,000
  cho 1,000,000,000,000 = 1.0*10^12

  5000 cho nichi = 5.0*10^15 nichi
  
  k = max 10^18 moji me
*/

#define DAY 5000000000000000
//#define DAY 3
#define ULL unsigned long long

int length(char S[101]){
  int i=0;
  while(1){
    if(S[i]=='\0'){
      break;
    }
    i++;
  }
  
  return i;
}

ULL power(ULL a, ULL b){
  ULL y = 1;
  for(ULL i=0; i<b; i++){
    y *= a;
  }
  return y;
}

int sn(char s){
  return s - 48;
}

int main(){  
  char S[101];
  unsigned long long k;
  scanf("%s", S);
  scanf("%lld", &k);

  int len = length(S);
 
  ULL a = 0;
  int ans;
  for(int i=0; i<len; i++){
    if(sn(S[i]) != 1){
      if(i+1 < k) ans = 1;
      else ans = sn(S[i]);
      break;
    }
  }

  printf("%d\n", ans);
  
  
  return 0;
}