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>
#include <stdlib.h>
#include <string.h>
#define DAY 5000000000000000

typedef struct{
  int num;
  int end;
} numRange;

void convert(char *str, int n, numRange arr[]){
  int i;
  for(i = 0; i < n; i++){
    arr[i].num = str[i] - '0';
    arr[i].end = i;
  }
}

void change_day(numRange arr[], int n){
  int i;
  for(i = 0; i < n; i++){
    if(i == 0) arr[i].end = (arr[i].end + 1) * arr[i].num - 1;
    else arr[i].end = arr[i-1].end + (arr[i].end - arr[i-1].end) * arr[i].num;
  }
} 

int answer(numRange arr[], int n, int k){
  int i;
  for(i = 0; i < n; i++){
    if(arr[i].end >= k-1)
      return arr[i].num;
  }
}

int main(void){
  long long i; int j;
  char *S; int n;
  int K, k = K;
  int ans;
  
  S = (char *)malloc(sizeof(char)*101);
  scanf("%s\n%d", S, &K);
  n = strlen(S);
  
  numRange arr[n];
  convert(S, n, arr);
  
  while(arr[0].num == 1){
    if(n == 0){
      printf("%d", 1);
      return 0;
    }
    for(j = 0; j < n-1; j++){
      arr[j] = arr[j+1];
      arr[j].end--;
    }
    n--; k--;
  }
  
  for(i = 0; i < DAY; i++){
    if(arr[0].end >= k-1){
      printf("%d", arr[0].num);
      return 0;
    }
    if(arr[n-2].end >= k-1){
      arr[n-2].end = k-1;
      n--;
    }
    change_day(arr, n);
  }
  
  ans = answer(arr, n, k);
  
  printf("%d", ans);
  return 0;
}