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 `x`

character after ^{th}`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 `s`

* after replacing all digits. It is guaranteed that *

`shift(s[i-1], s[i])`

`'z'`

.

**Example 1:**

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'

**Example 2:**

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'

**Constraints:**

`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`

.

class Solution {
public String replaceDigits(String s) {
}
}