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 OCaml (4.02.3) (judge verdict: Runtime Error)

open Scanf
open Printf

let tail s = String.sub s 1 (String.length s - 1)
let head s = int_of_string (String.sub s 0 1)

let n = float_of_int @@ 5000 * 10000 * 10000 * 10000

let pow i n = if i = 1
  then 1
  else begin
    let rec pow' k n = if n = 1 then k else pow' (k * i) (n - 1)
    in pow' i n
  end

let rec f s k = let i = head s in
  let a = log @@ float_of_int i in
  if k < n *. a
  then i
  else f (tail s) (k +. log (1. -. exp (n *. a -. k)))

let () = let ans = scanf "%s\n%d\n" (fun s k -> f s (log @@ float_of_int k)) in
  printf "%d\n" ans