Finish the solution so that it takes an input `n` (integer) and returns a string that is the decimal representation of the number grouped by commas after every 3 digits.

Assume: `0 <= n < 2147483647`

## Examples

```
       1  ->           "1"
      10  ->          "10"
     100  ->         "100"
    1000  ->       "1,000"
   10000  ->      "10,000"
  100000  ->     "100,000"
 1000000  ->   "1,000,000"
35235235  ->  "35,235,235"
```

Difficulty: EASY
Starter code: def group_by_commas(n):
	

Solutions:
def group_by_commas(n):
	return '{:,}'.format(n)


group_by_commas = '{:,}'.format


def group_by_commas(n):
	return f'{n:,}'


group_by_commas = lambda n: '{:,}'.format(n)


def group_by_commas(n):
	n = [i for i in str(n)]
	for i in range(len(n) - 4, -1, -3):
		n.insert(i + 1, ',')
	return ''.join(n)


def group_by_commas(n):
	if n < 0:
		if n >= -999:
			return str(n)
	elif n <= 999:
		return str(n)
	return format(n, ',')


def group_by_commas(n):
	rev = str(n)[::-1]
	groups = [rev[i:i + 3] for i in range(0, len(rev), 3)]
	return ','.join(groups)[::-1]


def group_by_commas(price):
	return '{:,}'.format(price)


import locale

def group_by_commas(n):
	return format(n, ',d')