String Decryption – Character Position

String Decryption – Character Position: The program must accept a string S (encrypted string) as the input. The encryption algorithm used to form the string S is given below.
– Each character in the original string is combined with its position on the left or right side.
– After combining all the characters with their positions in the original string, they are shuffled and separated by a hyphen.
The program must find the original string by decrypting the given string S. Then the program must print the original string as the output.

Boundary Condition(s):
5 <= Length of S <= 100

Input Format:
The first line contains S.

Output Format:
The first line contains the original string.

Example Input/Output 1:
Input:
4l-9k-R6-7a-k2-3i-S1-l5-8c

Output:
SkillRack

Explanation:
The encrypted string is 4l-9k-R6-7a-k2-3i-S1-l5-8c.
The position of l is 4.
The position of k is 9.
The position of R is 6.
The position of a is 7.
The position of k is 2.
The position of i is 3.
The position of S is 1.
The position of l is 5.
The position of c is 8.
After decrypting the given string, the original string is formed as SkillRack.
So SkillRack is printed as the output.

Example Input/Output 2:
Input:
6n-4t-1C-o2-d10-11y-t3-8a-o5-7C-n9

Output:
CottonCandy

a=list(map(str,input().split("-")))
s,b=[],[]
for i in a:
    x,y='',''
    for j in i:
        if j.isalpha():
            x+=j
        else:
            y+=j
    s.append(x)
    b.append(int(y))
for i in range(1,max(b)+1):
    for j in range(len(b)):
        if i==b[j]:
            print(s[j],end="")
    

a=[‘6n’,’4t’,’1C’,’o2′,’d10′,’11y’,’t3′,’8a’,’o5′,’7C’,’n9′] s,b=[],[] for i in a: x,y=”,” for j in i: if j.isalpha(): x+=j else: y+=j s.append(x) b.append(int(y)) for i in range(1,max(b)+1): for j in range(len(b)): if i==b[j]: print(s[j],end=””)