Obfuscation S47315

Obfuscation of a string $s = c_1 c_2 c_3 \dots c_k$ with respect to a natural number $n$ consists of changing all the letters $c_i \in s$ by the letter that is $n$ places after $c_i$ in the alphabet.

For example, ofuscacio(1,'A') = 'B', because 'B' is one place after 'A'. Other examples: ofuscacio(4,'B') = 'F' and ofuscacio(2,'AB') = 'CD'.

The function recursive void ofuscacio(int n, string s) must be implemented with the following specification:

PRE: The input is an integer $n$ such that $0 \leq n \leq 20$ and a string of characters $s = c_1 c_2 c_3 \dots c_k$ such that $k > 0$ and $\forall c_i \in s, 'A' \leq c_i \leq 'F'$.

POST: writes to the output channel cout the obfuscation of the string of characters $s$ with respect to $n$.

Observation

If necessary, you can use the method pop_back() for vectors and string.

Only recursive solutions are accepted.

Just send the function. The rest will be ignored.

Input

An integer $n$ such that $0 \leq n \leq 20$ and a string of characters $s = c_1 c_2 c_3 \dots c_k$ such that $k > 0$ and $\forall c_i \in s, 'A' \leq c_i \leq 'F'$.

Output

For each pair n, s, the obfuscation of s with respect to n.