# Task

There are `n` bears in the orchard and they picked a lot of apples. 

They distribute apples like this:

```
The first bear divided the apple into n piles, each with the same number. leaving an apple at last, and the apple was thrown away. Then he took 1 pile and left the n-1 pile.

The second bear divided the apple into n piles, each with the same number. leaving an apple at last, and the apple was thrown away. Then he took 1 pile and left the n-1 pile.

and so on..
```

Hmm.. your task is coming, please calculate the minimum possible number of these apples (before they start distributing)

# Input/Output


`[input]` integer `n`

The number of bears.

`2 <= n <= 9` (JS, Crystal and C)
`2 <= n <= 50` (Ruby and Python)

`[output]` an integer

The minimum possible number of apples.

# Example

For `n = 5`, the output should be `3121`.

```
5 bears distribute apples:

1st bear divided the apples into 5 piles, each pile has 624 apples, and 1 apple was thrown away.
3121 - 1 = 624 x 5
1st bear took 1 pile of apples, 2496 apples left.
3121 - 1 - 624 = 2496

2nd bear divided the apples into 5 piles, each pile has 499 apples, and 1 apple was thrown away.
2496 - 1 = 499 x 5
2nd bear took 1 pile of apples, 1996 apples left.
2496 - 1 - 499 = 1996

3rd bear divided the apples into 5 piles, each pile has 399 apples, and 1 apple was thrown away.
1996 - 1 = 399 x 5
3rd bear took 1 pile of apples, 1596 apples left.
1996 - 1 - 399 = 1596

4th bear divided the apples into 5 piles, each pile has 319 apples, and 1 apple was thrown away.
1596 - 1 = 319 x 5
4th bear took 1 pile of apples, 1276 apples left.
1596 - 1 - 319 = 1276

5th bear divided the apples into 5 piles, each pile has 255 apples, and 1 apple was thrown away.
1276 - 1 = 255 x 5
5th bear took 1 pile of apples, 1020 apples left.
1276 - 1 - 255 = 1020
```

Difficulty: introductory

Starter code:

def how_many_apples(n):
	

Solutions:
how_many_apples=lambda n:n**n-(n-1)+[0,4][n==2]

def how_many_apples(n):
    return n ** n - n + 1 if n > 2 else 7

def how_many_apples(n):
    if n == 2:
        return 7
    else:
        return n**n-(n-1)

def how_many_apples(n):
    if n==2:
        return 7
    return (n**(n-1))*n-n+1

def how_many_apples(n):
    return n**n-(n-1) if n!=2 else 7

def how_many_apples(n):
    if n == 2:
        return 7
    if n == 3:
        return 25
    x = (how_many_apples(n-1) + (n-3)) * (n-1)
    for i in range(n):
        x = x * n // (n-1) + 1
    return x

def how_many_apples(n):
    if n == 2:
        return 7
    if n == 3:
        return 25
    x = how_many_apples(n-1) * (n-1) + (n-1) * (n-3)
    for i in range(n):
        x = x * n // (n-1) + 1
    return x

def how_many_apples(n):
    # Monkey and Coconut Problem
    # https://www.youtube.com/watch?v=U9qU20VmvaU
    return n ** max(n, 3) - n + 1

def how_many_apples(n):
    return n**max(n,3)-n+1

how_many_apples=lambda n:n>2and n**n-n+1or 7