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# Move the 1 to position 0. You can do this in one step with .pop() and .insert().

helppppppppppppppppppppppppppp

lists.py
```the_list = ["a", 2, 3, 1, False, [1, 2, 3]]

# Your code goes below here
```

No need for the temp variable. It can be done with pop() and .insert() on a single line.

the_list.insert(0, the_list.pop(3))

MOD

From the Lists Redux manipulating lists Challenge:

Move the 1 to position 0. You can do this in one step with .pop() and .insert().

Given the list:

```the_list = ["a", 2, 3, 1, False, [1, 2, 3]]
```

The "1" is the 4-th item in the list and can be reference as `list[3]` (using 0-based counting).

Use the `pop()` method to "pop" an item off of the list. Assign this to a temp variable:

```temp = the_list.pop(3)
```

Use the `insert()` method to "insert" an item into a list at a specific position. Inserting `temp` into `the_list` at position 0::

```the_list.insert(0, temp)
```

All together, running in ipython, it would look like:

```In [84]: the_list = ["a", 2, 3, 1, False, [1, 2, 3]]

In [85]: the_list[3]
Out[85]: 1

In [86]: temp = the_list.pop(3)

In [87]: temp
Out[87]: 1

In [88]: the_list
Out[88]: ['a', 2, 3, False, [1, 2, 3]]

In [89]: the_list.insert(0, temp)

In [90]: the_list
Out[90]: [1, 'a', 2, 3, False, [1, 2, 3]]
```