You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

• For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.

For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).

Return safter replacing all digits. It is guaranteed thatshift(s[i-1], s[i])will never exceed'z'.

Input: s = "a1c1e1" Output: "abcdef" Explanation: The digits are replaced as follows: - s[1] -> shift('a',1) = 'b' - s[3] -> shift('c',1) = 'd' - s[5] -> shift('e',1) = 'f' 

Input: s = "a1b2c3d4e" Output: "abbdcfdhe" Explanation: The digits are replaced as follows: - s[1] -> shift('a',1) = 'b' - s[3] -> shift('b',2) = 'd' - s[5] -> shift('c',3) = 'f' - s[7] -> shift('d',4) = 'h' 

• 1 <= s.length <= 100
• s consists only of lowercase English letters and digits.
• shift(s[i-1], s[i]) <= 'z' for all odd indices i.

implSolution{pubfnreplace_digits(s:String) -> String{letmut s = s.into_bytes();for i in(1..s.len()).step_by(2){ s[i] += s[i - 1] - b'0';}String::from_utf8(s).unwrap()}}