A beauty factor.py

# Simple Brute Force implementation 
#First we check that whether the length (k) is 9 or not because if k ==9  then we can have only 1 number that is 123456789 becz number must be distinct and greater than zero.
# Step 2 : If start with the smallest number and remove all zero and indentical number (1231 here 1 is repeated two times which should not be as each number must be distinct) and
# check there beautyFactor . if they match print it else increment the number and recursivly check it len of the number is equal to k.


def checkBeautyFactor(n):
    su = 0
    for i in str(n):
        su+= int(i)
        
    if len(str(su))>1:
        su = checkBeautyFactor(str(su))
        
    return su

b,k = map(int,input().split())


if k == 9:
    if b==9:
        print(123456789)
    else:
        print(-1)
else:
    n = "1" + "0"*(k-1)
    while True:
        if "0" in str(n):
            n = int(n)
            n+=1
            continue
        n = str(n)
        if len(set(n)) != len(n):
            n = int(n)
            n+=1
            continue
            
        beautyFactor = checkBeautyFactor(str(n))
        
        if beautyFactor == b:
            print(n)
            break
            
        n = int(n)
        n+=1
        if len(str(n))>k:
            print(-1)
            break