meetforever/GRPO_CL huggingface dataset:
user: We have an array with string digits that occurrs more than once, for example, ```arr = ['1', '2', '2', '2', '3', '3']```. How many different string numbers can be generated taking the 6 elements at a time?

We present the list of them below in an unsorted way:
``` 
['223213', '312322', '223312', '222133', '312223', '223321', '223231', '132223', '132322', '223132', '322321', '322312', '231322', '222313', '221233', '213322', '122323', '321322', '221332', '133222', '123232', '323221', '222331', '132232', '321232', '212323', '232213', '232132', '331222', '232312', '332212', '213223', '123322', '322231', '321223', '231232', '233221', '231223', '213232', '312232', '233122', '322132', '223123', '322123', '232231', '323122', '323212', '122233', '212233', '123223', '332221', '122332', '221323', '332122', '232123', '212332', '232321', '322213', '233212', '313222']
``` 
There are ```60``` different numbers and ```122233``` is the lowest one and ```332221``` the highest of all of them.

Given an array, ```arr```, with string digits (from '0' to '9'), you should give the exact amount of different numbers that may be generated with the lowest and highest value but both converted into integer values, using all the digits given in the array for each generated string number.

The function will be called as ```proc_arr()```.
```python
proc_arr(['1', '2', '2', '3', '2', '3']) == [60, 122233, 332221]
```

If the digit '0' is present in the given array will produce string numbers with leading zeroes, that will not be not taken in account when they are converted to integers.
```python
proc_arr(['1','2','3','0','5','1','1','3']) == [3360, 1112335, 53321110]
```
You will never receive an array with only one digit repeated n times.

Features of the random tests:
```
Low performance tests:
Number of tests: 100
Arrays of length between 6 and 10

Higher performance tests:
Number of tests: 100
Arrays of length between 30 and 100
```

Answer: {"fn_name": "proc_arr", "inputs": [[["1", "2", "2", "3", "2", "3"]]], "outputs": [[[60, 122233, 332221]]]}
user: Welcome young Jedi! In this Kata you must create a function that takes an amount of US currency in `cents`, and returns a dictionary/hash which shows the least amount of coins used to make up that amount. The only coin denominations considered in this exercise are: `Pennies (1¢), Nickels (5¢), Dimes (10¢) and Quarters (25¢)`.
Therefor the dictionary returned should contain exactly 4 key/value pairs.

Notes:

* If the function is passed either 0 or a negative number, the function should return the dictionary with all values equal to 0.
* If a float is passed into the function, its value should be be rounded down, and the resulting dictionary should never contain fractions of a coin.


## Examples
```
loose_change(56)    ==>  {'Nickels': 1, 'Pennies': 1, 'Dimes': 0, 'Quarters': 2}
loose_change(-435)  ==>  {'Nickels': 0, 'Pennies': 0, 'Dimes': 0, 'Quarters': 0}
loose_change(4.935) ==>  {'Nickels': 0, 'Pennies': 4, 'Dimes': 0, 'Quarters': 0}
```

Answer: {"fn_name": "loose_change", "inputs": [[56], [0], [100], [-3], [7.9]], "outputs": [[{"Nickels": 1, "Pennies": 1, "Dimes": 0, "Quarters": 2}], [{"Nickels": 0, "Pennies": 0, "Dimes": 0, "Quarters": 0}], [{"Nickels": 0, "Pennies": 0, "Dimes": 0, "Quarters": 4}], [{"Nickels": 0, "Pennies": 0, "Dimes": 0, "Quarters": 0}], [{"Nickels": 1, "Pennies": 2, "Dimes": 0, "Quarters": 0}]]}
user: Consider the prime number `23`. If we sum the square of its digits we get:
`2^2 + 3^2 = 13`, then for `13: 1^2 + 3^2 = 10`, and finally for `10: 1^2 + 0^2 = 1`. 

Similarly, if we start with prime number `7`, the sequence is: `7->49->97->130->10->1`.

Given a range, how many primes within that range will eventually end up being `1`? 

The upperbound for the range is `50,000`. A range of `(2,25)` means that: `2 <= n < 25`. 

Good luck!

If you like this Kata, please try:

[Prime reversion](https://www.codewars.com/kata/59b46276afcda204ed000094)

[Domainant primes](https://www.codewars.com/kata/59ce11ea9f0cbc8a390000ed)

Answer: {"fn_name": "solve", "inputs": [[1, 25], [100, 1000], [100, 2000], [100, 3000], [100, 4000]], "outputs": [[4], [28], [47], [65], [95]]}
user: This is a follow-up from my previous Kata which can be found here: http://www.codewars.com/kata/5476f4ca03810c0fc0000098

This time, for any given linear sequence, calculate the function [f(x)] and return it as a function in Javascript or Lambda/Block in Ruby.

For example:

```python
get_function([0, 1, 2, 3, 4])(5) => 5
get_function([0, 3, 6, 9, 12])(10) => 30
get_function([1, 4, 7, 10, 13])(20) => 61
```

Assumptions for this kata are:
```
The sequence argument will always contain 5 values equal to f(0) - f(4).
The function will always be in the format "nx +/- m", 'x +/- m', 'nx', 'x' or 'm'
If a non-linear sequence simply return 'Non-linear sequence' for javascript, ruby, and python. For C#, throw an ArgumentException.
```

Answer: {"fn_name": "get_function", "inputs": [[[1, 2, 4, 7, 11]], [[0, 1, 2, 3, 100000]], [[0, 1, 2, 100000, 4]], [[0, 1, 100000, 3, 4]], [[0, 100000, 2, 3, 4]]], "outputs": [["Non-linear sequence"], ["Non-linear sequence"], ["Non-linear sequence"], ["Non-linear sequence"], ["Non-linear sequence"]]}
user: Perhaps it would be convenient to solve first ```Probabilities for Sums in Rolling Cubic Dice``` see at: http://www.codewars.com/kata/probabilities-for-sums-in-rolling-cubic-dice

Suppose that we roll dice that we never used, a tetrahedral die that has only 4 sides(values 1 to 4), or perhaps better the dodecahedral one with 12 sides and 12 values (from 1 to 12), or look! an icosahedral die with 20 sides and 20 values (from 1 to 20).

Let's roll dice of same type and try to see the distribution of the probability for all the possible sum values.

Suppose that we roll 3 tetrahedral dice:
```
Sum   Prob    Amount of hits           Combinations
3   0.015625     1             {1,1,1}
4   0.046875     3             {1,2,1},{2,1,1},{1,1,2}
5   0.09375      6             {3,1,1},{1,1,3},{1,2,2},{1,3,1},{2,2,1},{2,1,2}
6   0.15625     10             {2,2,2},{1,3,2},{3,2,1},{1,4,1},{2,1,3},{1,1,4},{4,1,1},{3,1,2},{1,2,3},{2,3,1}
7   0.1875      12             {2,3,2},{1,2,4},{2,1,4},{1,4,2},{3,2,2}, {4,2,1},{3,3,1},{3,1,3},{2,2,3},{4,1,2},{2,4,1},{1,3,3}
8   0.1875      12             {2,3,3},{3,3,2},{2,2,4},{4,3,1},{2,4,2},{4,1,3},{4,2,2},{3,4,1},{1,3,4},{3,2,3},{3,1,4},{1,4,3}
9   0.15625     10             {3,3,3},{4,4,1},{2,4,3},{4,2,3},{4,3,2},{1,4,4},{4,1,4},{3,4,2},{2,3,4},{3,2,4}
10  0.09375      6             {3,4,3},{4,4,2},{3,3,4},{4,2,4},(2,4,4),{4,3,3}
11  0.046875     3             {4,3,4},{3,4,4},{4,4,3}
12  0.015625     1             {4,4,4}
tot: 1.0000
```
Note that the total sum of all the probabilities for each end has to be ```1.0000```

The register of hits per each sum value will be as follows:
```
[[3, 1], [4, 3], [5, 6], [6, 10], [7, 12], [8, 12], [9, 10], [10, 6], [11, 3], [12, 1]]
```
Create the function ```reg_sum_hits()``` that receive two arguments:

- number of dice, ```n```
- number of sides of the die, ```s```

The function will output a list of pairs with the sum value and the corresponding hits for it.

For the example given above:
```python
reg_sum_hits(3, 4) == [[3, 1], [4, 3], [5, 6], [6, 10], [7, 12], [8, 12], [9, 10], [10, 6], [11, 3], [12, 1]]
```
You will find more examples in the Example Test Cases.

Assumption: If the die has n sides, the values of it are from ```1``` to ```n```

Answer: {"fn_name": "reg_sum_hits", "inputs": [[3, 4], [4, 4], [3, 6], [4, 6]], "outputs": [[[[3, 1], [4, 3], [5, 6], [6, 10], [7, 12], [8, 12], [9, 10], [10, 6], [11, 3], [12, 1]]], [[[4, 1], [5, 4], [6, 10], [7, 20], [8, 31], [9, 40], [10, 44], [11, 40], [12, 31], [13, 20], [14, 10], [15, 4], [16, 1]]], [[[3, 1], [4, 3], [5, 6], [6, 10], [7, 15], [8, 21], [9, 25], [10, 27], [11, 27], [12, 25], [13, 21], [14, 15], [15, 10], [16, 6], [17, 3], [18, 1]]], [[[4, 1], [5, 4], [6, 10], [7, 20], [8, 35], [9, 56], [10, 80], [11, 104], [12, 125], [13, 140], [14, 146], [15, 140], [16, 125], [17, 104], [18, 80], [19, 56], [20, 35], [21, 20], [22, 10], [23, 4], [24, 1]]]]}
user: Every number may be factored in prime factors.

For example, the number 18 may be factored by its prime factors ``` 2 ``` and ```3```
```
18 = 2 . 3 . 3 = 2 . 3²
```
The sum of the prime factors of 18 is  ```2 + 3 + 3 = 8```

But some numbers like 70 are divisible by the sum of its prime factors:
```
70 = 2 . 5 . 7 # sum of prime factors = 2 + 5 + 7 = 14
and 70 is a multiple of 14
```
Of course that primes would fulfill this property, but is obvious, because the prime decomposition of a number, is the number itself and every number is divisible by iself. That is why we will discard every prime number in the results

We are interested in collect the integer positive numbers (non primes) that have this property in a certain range ```[a, b]``` (inclusive).

Make the function ```mult_primefactor_sum()```, that receives the values ```a```, ```b``` as limits of the range ```[a, b]``` and ```a < b``` and outputs the sorted list of these numbers.

Let's see some cases:
```python
mult_primefactor_sum(10, 100) == [16, 27, 30, 60, 70, 72, 84] 

mult_primefactor_sum(1, 60) == [1, 4, 16, 27, 30, 60]
```

Answer: {"fn_name": "mult_primefactor_sum", "inputs": [[10, 100], [80, 150], [90, 200]], "outputs": [[[16, 27, 30, 60, 70, 72, 84]], [[84, 105, 150]], [[105, 150, 180]]]}
user: Implement a class/function, which should parse time expressed as `HH:MM:SS`, or `null/nil/None` otherwise.

Any extra characters, or minutes/seconds higher than 59 make the input invalid, and so should return `null/nil/None`.

Answer: {"fn_name": "to_seconds", "inputs": [["00:00:00"], ["01:02:03"], ["01:02:60"], ["01:60:03"], ["99:59:59"], ["0:00:00"], ["00:0:00"], ["00:00:0"], ["00:00:00\n"], ["\n00:00:00"], ["100:59:59"], ["09:059:59"], ["09:159:59"], ["09:59:059"], ["09:59:159"]], "outputs": [[0], [3723], [null], [null], [359999], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null]]}
user: Welcome

This kata is inspired by This Kata


We have a string s

We have a number n

Here is a function that takes your string, concatenates the even-indexed chars to the front, odd-indexed chars to the back.

Examples

    s = "Wow Example!"
    result = "WwEapeo xml!"
    s = "I'm JomoPipi"
    result = "ImJm ii' ooPp"
    
The Task:

return the result of the string after applying the function to it n times.

example where s = "qwertyuio" and n = 2:

    after 1 iteration  s = "qetuowryi"
    after 2 iterations s = "qtorieuwy"
    return "qtorieuwy"

 Note 

there's a lot of tests, big strings,
and n is greater than a billion

so be ready to optimize.

after doing this: do it's best friend!

# Check out my other kata!

        
Matrix Diagonal Sort OMG
String -> N iterations -> String
String -> X iterations -> String
ANTISTRING
Array - squareUp b!
Matrix - squareUp b!
Infinitely Nested Radical Expressions
pipi Numbers!

Answer: {"fn_name": "jumbled_string", "inputs": [["Such Wow!", 1], ["better example", 2], ["qwertyuio", 2], ["Greetings", 8], ["I like it!", 1234], ["codingisfornerdsyounerd", 10101010], ["this_test_will_hurt_you", 12345678987654321]], "outputs": [["Sc o!uhWw"], ["bexltept merae"], ["qtorieuwy"], ["Gtsegenri"], ["Iiei t kl!"], ["criyinodedsgufrnodnoser"], ["ti_etwl_utyuhsts_ilhr_o"]]}
user: In the previous Kata we discussed the OR case.

We will now discuss the AND case, where rather than calculating the probablility for either of two (or more) possible results, we will calculate the probability of receiving all of the viewed outcomes.

For example, if we want to know the probability of receiving head OR tails in two tosses of a coin, as in the last Kata we add the two probabilities together. However if we want to know the probability of receiving head AND tails, in that order, we have to work differently.

The probability of an AND event is calculated by the following rule:

`P(A ∩ B) = P(A | B) * P(B)`

or

`P(B ∩ A) = P(B | A) * P(A)`

That is, the probability of A and B both occuring is equal to the probability of A given B occuring times the probability of B occuring or vice versa. 
If the events are mutually exclusive like in the case of tossing a coin, the probability of A occuring if B has occured is equal to the probability of A occuring by itself. In this case, the probability can be written as the below:

`P(A ∩ B) = P(A) * P(B)`

or

`P(B ∩ A) = P(B) * P(A)`

Applying to the heads and tails case:

`P(H ∩ T) = P(0.5) * P(0.5)`

or

`P(H ∩ T) = P(0.5) * P(0.5)`

The task:

You are given a random bag of 10 balls containing 4 colours. `Red`, `Green`, `Yellow` and `Blue`. You will also be given a sequence of 2 balls of any colour e.g. `Green` and `Red` or `Yellow` and `Yellow`. 

You have to return the probability of pulling a ball at random out of the bag and it matching the first colour and then pulling a ball at random out of the bag and it matching the second colour.

You will be given a boolean value of `true` or `false` which indicates whether the balls that is taken out in the first draw is replaced in to the bag for the second draw. Hint: this will determine whether the events are mutually exclusive or not.

You will receive two arrays and a boolean value. The first array contains the colours of the balls in the bag and the second contains the colour of the two balls you have to receive. As per above the final boolean value will indicate whether the bals arel being replaced `true` or not `false`.

Return the probability to 3 decimal places.

e.g. `[["y","r","g","b","b","y","r","g","r","r"],["r","b"],false]`


Other Kata in this series:
Statistics in Kata 1: OR case - Unfair dice

Answer: {"fn_name": "ball_probability", "inputs": [[[["red", "blue", "yellow", "green", "red", "blue", "yellow", "green", "red", "blue"], ["red", "blue"], true]], [[["red", "blue", "yellow", "green", "red", "blue", "yellow", "green", "red", "blue"], ["red", "red"], true]], [[["red", "red", "yellow", "green", "red", "red", "yellow", "green", "red", "red"], ["blue", "blue"], true]], [[["red", "blue", "yellow", "green", "red", "blue", "yellow", "green", "red", "blue"], ["red", "blue"], false]], [[["red", "blue", "yellow", "green", "red", "blue", "yellow", "green", "red", "blue"], ["red", "red"], false]]], "outputs": [[0.09], [0.09], [0], [0.1], [0.067]]}
user: # Task

* **_Given_** *three integers* `a` ,`b` ,`c`, **_return_** *the **_largest number_** obtained after inserting the following operators and brackets*: `+`, `*`, `()`
* In other words , **_try every combination of a,b,c with [*+()] , and return the Maximum Obtained_**
___
# Consider an Example :

**_With the numbers are 1, 2 and 3_** , *here are some ways of placing signs and brackets*:

* `1 * (2 + 3) = 5`
* `1 * 2 * 3 = 6`
* `1 + 2 * 3 = 7`
* `(1 + 2) * 3 = 9`

So **_the maximum value_** that you can obtain is  **_9_**.

___
# Notes 

* **_The numbers_** *are always* **_positive_**. 
* **_The numbers_** *are in the range* **_(1  ≤  a, b, c  ≤  10)_**.
* *You can use the same operation* **_more than once_**.
* **It's not necessary** *to place all the signs and brackets*.
* **_Repetition_** *in numbers may occur* .
* You **_cannot swap the operands_**. For instance, in the given example **_you cannot get expression_** `(1 + 3) * 2 = 8`.

___
# Input >> Output Examples:

```
expressionsMatter(1,2,3)  ==>  return 9
```
## **_Explanation_**:
*After placing signs and brackets, the **_Maximum value_** obtained from the expression* `(1+2) * 3 = 9`.
___

```
expressionsMatter(1,1,1)  ==>  return 3
```
## **_Explanation_**:
*After placing signs, the **_Maximum value_** obtained from the expression is* `1 + 1 + 1 = 3`.
___

```
expressionsMatter(9,1,1)  ==>  return 18
```
## **_Explanation_**:
*After placing signs and brackets, the **_Maximum value_** obtained from the expression is* `9 * (1+1) = 18`.
___
___
___

# [Playing with Numbers Series](https://www.codewars.com/collections/playing-with-numbers)

# [Playing With Lists/Arrays Series](https://www.codewars.com/collections/playing-with-lists-slash-arrays)

# [Bizarre Sorting-katas](https://www.codewars.com/collections/bizarre-sorting-katas)

# [For More Enjoyable Katas](http://www.codewars.com/users/MrZizoScream/authored)
___

## ALL translations are welcomed

## Enjoy Learning !!
# Zizou

Answer: {"fn_name": "expression_matter", "inputs": [[2, 1, 2], [2, 1, 1], [1, 1, 1], [1, 2, 3], [1, 3, 1], [2, 2, 2], [5, 1, 3], [3, 5, 7], [5, 6, 1], [1, 6, 1], [2, 6, 1], [6, 7, 1], [2, 10, 3], [1, 8, 3], [9, 7, 2], [1, 1, 10], [9, 1, 1], [10, 5, 6], [1, 10, 1]], "outputs": [[6], [4], [3], [9], [5], [8], [20], [105], [35], [8], [14], [48], [60], [27], [126], [20], [18], [300], [12]]}
user: There's a **"3 for 2"** (or **"2+1"** if you like) offer on mangoes. For a given quantity and price (per mango), calculate the total cost of the mangoes.

### Examples
```python
mango(3, 3) ==> 6    # 2 mangoes for 3 = 6; +1 mango for free
mango(9, 5) ==> 30   # 6 mangoes for 5 = 30; +3 mangoes for free
```

Answer: {"fn_name": "mango", "inputs": [[3, 3], [9, 5]], "outputs": [[6], [30]]}
user: Your task it to return ```true``` if the fractional part (rounded to 1 digit) of the result (```a``` / ```b```) exists, more than ```0``` and is multiple of ```n```.
Otherwise return ```false```. (For Python return True or False)

All arguments are positive digital numbers.

Rounding works like toFixed() method. (if more than...5 rounds up)

Find exapmles below: 

```

isMultiple(5, 2, 3) -> false // 2.5 -> 5 is not multiple of 3
isMultiple(5, 3, 4) -> false // 1.7 -> 7 is not multiple of 4
isMultiple(5, 4, 3) -> true // 1.3 -> 3 is multiple of 3
isMultiple(666, 665, 2) -> false // 1.0 -> return false


```

Answer: {"fn_name": "isMultiple", "inputs": [[5, 2, 3], [5, 3, 4], [5, 4, 3], [666, 665, 2], [3691401, 1892272, 5]], "outputs": [[false], [false], [true], [false], [false]]}
user: # Task

Given an array of integers, sum consecutive even numbers and consecutive odd numbers. Repeat the process while it can be done and return the length of the final array.

# Example

For `arr = [2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9]`
 
The result should be `6`.

```
[2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9]  -->
         2+2+6       0+2+0     5+5+7+7       3+3+9
[2, 1,   10,    5,    2,        24,     4,   15   ] -->
                               2+24+4
[2, 1,   10,    5,             30,           15   ]
The length of final array is 6
```

# Input/Output

- `[input]` integer array `arr`

  A non-empty array, 

  `1 ≤ arr.length ≤ 1000`

  `0 ≤ arr[i] ≤ 1000`


- `[output]` an integer

  The length of the final array

Answer: {"fn_name": "sum_groups", "inputs": [[[2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9]], [[2, 1, 2, 2, 6, 5, 0, 2, 0, 3, 3, 3, 9, 2]], [[2]], [[1, 2]], [[1, 1, 2, 2]]], "outputs": [[6], [5], [1], [2], [1]]}
user: In this kata you're expected to sort an array of 32-bit integers in ascending order of the number of **on** bits they have.

E.g Given the array **[7, 6, 15, 8]**

- 7 has **3 on** bits  (000...0**111**)
- 6 has **2 on** bits  (000...00**11**)
- 15 has **4 on** bits (000...**1111**)
- 8 has **1 on** bit   (000...**1**000)

So the array in sorted order would be **[8, 6, 7, 15]**.

In cases where two numbers have the same number of bits, compare their real values instead.

E.g between 10 **(...1010)** and 12 **(...1100)**, they both have the same number of **on** bits '**2**' but the integer 10 is less than 12 so it comes first in sorted order.

Your task is to write the function `sortBybit()` that takes an array of integers and sort them as described above.

```if-not:haskell
Note: Your function should modify the input rather than creating a new array.
```

Answer: {"fn_name": "sort_by_bit", "inputs": [[[3, 8, 3, 6, 5, 7, 9, 1]], [[9, 4, 5, 3, 5, 7, 2, 56, 8, 2, 6, 8, 0]]], "outputs": [[[1, 8, 3, 3, 5, 6, 9, 7]], [[0, 2, 2, 4, 8, 8, 3, 5, 5, 6, 9, 7, 56]]]}
user: A palindrome is a series of characters that read the same forwards as backwards such as "hannah", "racecar" and "lol".

For this Kata you need to write a function that takes a string of characters and returns the length, as an integer value, of longest alphanumeric palindrome that could be made by combining the characters in any order but using each character only once. The function should not be case sensitive.

For example if passed "Hannah" it should return 6 and if passed "aabbcc_yYx_" it should return 9 because one possible palindrome would be "abcyxycba".

Answer: {"fn_name": "longest_palindrome", "inputs": [["A"], ["Hannah"], ["xyz__a_/b0110//a_zyx"], ["$aaabbbccddd_!jJpqlQx_.///yYabababhii_"], [""]], "outputs": [[1], [6], [13], [25], [0]]}
user: Write a function 

```python
vowel_2_index```

that takes in a string and replaces all the vowels [a,e,i,o,u] with their respective positions within that string. 
E.g: 

```python
vowel_2_index('this is my string') == 'th3s 6s my str15ng'
vowel_2_index('Codewars is the best site in the world') == 'C2d4w6rs 10s th15 b18st s23t25 27n th32 w35rld'
vowel_2_index('') == ''
```
 Your function should be case insensitive to the vowels.

Answer: {"fn_name": "vowel_2_index", "inputs": [["this is my string"], ["Codewars is the best site in the world"], ["Tomorrow is going to be raining"], [""], ["90's cornhole Austin, pickled butcher yr messenger bag gastropub next level leggings listicle meditation try-hard Vice. Taxidermy gastropub gentrify, meh fap organic ennui fingerstache pickled vegan. Seitan sustainable PBR&B cornhole VHS. Jean shorts actually bitters ugh blog Intelligentsia. Artisan Kickstarter DIY, fixie cliche salvia lo-fi four loko. PBR&B Odd Future ugh fingerstache cray Wes Anderson chia typewriter iPhone bespoke four loko, Intelligentsia photo booth direct trade. Aesthetic Tumblr Portland XOXO squid, synth viral listicle skateboard four dollar toast cornhole Blue Bottle semiotics."]], "outputs": [["th3s 6s my str15ng"], ["C2d4w6rs 10s th15 b18st s23t25 27n th32 w35rld"], ["T2m4rr7w 10s g1415ng t20 b23 r2627n29ng"], [""], ["90's c7rnh11l13 1516st19n, p24ckl28d b32tch36r yr m43ss46ng49r b53g g57str61p63b n67xt l72v74l l78gg81ngs l87st90cl93 m96d98t100t102103n try-h111rd V116c118. T122x124d126rmy g132str136p138b g142ntr146fy, m152h f156p 159rg162n164c 167nn170171 f174ng177rst181ch184 p187ckl191d v195g197n. S202203t205n s209st212213n215bl218 PBR&B c227rnh231l233 VHS. J241242n sh247rts 252ct255256lly b262tt265rs 269gh bl275g 278nt281ll284g286nts290291. 294rt297s299n K303ckst308rt311r D315Y, f320x322323 cl327ch330 s333lv336337 l340-f343 f346347r l351k353. PBR&B 362dd F367t369r371 373gh f378ng381rst385ch388 cr392y W396s 399nd402rs405n ch410411 typ416wr419t421r 424Ph427n429 b432sp435k437 f440441r l445k447, 450nt453ll456g458nts462463 ph467t469 b472473th d478r480ct tr486d488. 491492sth496t498c T502mblr P509rtl513nd X518X520 sq524525d, synth v536r538l l542st545cl548 sk552t554b556557rd f562563r d567ll570r t574575st c580rnh584l586 Bl590591 B594ttl598 s601m603604t606cs."]]}