Vb encryption function into the Delphi for help-CodePudding

 Attribute VB_Name="Module1" 
The Public Function Encode (ByVal s As String) As String 'encryption 
If Len (s)=0 Then the Exit Function 
Dim buff () As Byte 
Buff=StrConv (s, vbFromUnicode) 
Dim As Long I 
Dim j As Byte 
Dim k As Byte, m As Byte 
Dim MSTR As String 
'encrypted MSTR="ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz" contains characters 
Dim outs As String 
I=UBound (buff) + 1 
Outs=Space (2 * I) 
Dim temps As String 
For I=0 To UBound (buff) 
Randomize the Time 
J=CByte (5 * (math.h Rnd ()) + 0) 'biggest generated random number is only five, can't again big, big again, will use a byte 
Buff (I)=buff (I) Xor j 
K=buff (I) the Mod Len (MSTR) 
M=buff (I) \ Len (MSTR) 
M=m * 2 ^ 3 + j 
Temps=Mid (MSTR, k + 1, 1) + Mid (MSTR, m + 1, 1) 
Mid (outs, 2 * I + 1, 2)=temps 
Next 
Encode=outs 
End the Function 
The Public Function Decode (ByVal s As String) As String 'decryption 
On Error GoTo myERR 
Dim As Long I 
Dim j As Byte 
Dim k As Byte 
Dim m As Byte 
Dim MSTR As String 
MSTR="ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz" 
Dim t1 As String, t2 As String 
Dim buff () As Byte 
Dim n As Long 
N=0 
For I=1 To Len (s) Step 2 
T1=Mid (s, I, 1) 
T2=Mid (s, I + 1, 1) 
K=InStr (1, MSTR, t1) - 1 
M=InStr (1, MSTR, t2) - 1 
J=2 ^ 3 m \ 
M=m - j * 2 ^ 3 
ReDim Preserve buff (n) 
Buff (n)=j * Len (MSTR) + k 
Buff (n)=buff (n) Xor m 
N=n + 1 
Next 
Decode=StrConv (buff, vbUnicode) 
The Exit Function 
MyERR: 
Decode="" 
End the Function

Above is found online vb string encryption function, used in a small program, how do I add the vb decryption function into Delphi?
Only 20 points, all give up