There is a special typewriter with lowercase English letters `'a'` to `'z'` arranged in a **circle** with a **pointer**. A character can **only** be typed if the pointer is pointing to that character. The pointer is **initially** pointing to the character `'a'`.

Each second, you may perform one of the following operations:

*   Move the pointer one character **counterclockwise** or **clockwise**.
*   Type the character the pointer is **currently** on.

Given a string `word`, return the **minimum** number of seconds to type out the characters in `word`.

**Example 1:**

**Input:** word =  "abc "
**Output:** 5
**Explanation:** 
The characters are printed as follows:
- Type the character 'a' in 1 second since the pointer is initially on 'a'.
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer clockwise to 'c' in 1 second.
- Type the character 'c' in 1 second.

**Example 2:**

**Input:** word =  "bza "
**Output:** 7
**Explanation:**
The characters are printed as follows:
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer counterclockwise to 'z' in 2 seconds.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'a' in 1 second.
- Type the character 'a' in 1 second.

**Example 3:**

**Input:** word =  "zjpc "
**Output:** 34
**Explanation:**
The characters are printed as follows:
- Move the pointer counterclockwise to 'z' in 1 second.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'j' in 10 seconds.
- Type the character 'j' in 1 second.
- Move the pointer clockwise to 'p' in 6 seconds.
- Type the character 'p' in 1 second.
- Move the pointer counterclockwise to 'c' in 13 seconds.
- Type the character 'c' in 1 second.

**Constraints:**

*   `1 <= word.length <= 100`
*   `word` consists of lowercase English letters.


def minTimeToType(word: str) -> int:
    time = 0
    position = 0
    for c in word:
        next_position = ord(c) - ord('a')
        diff = abs(next_position - position)
        time += min(diff, 26 - diff) + 1
        position = next_position
    return time
