Kata

Some numbers have funny properties. For example:

• 89 --> 8¹ + 9² = 89 * 1
• 695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
• 46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

Given two positive integers n and p, we want to find a positive integer k, if it exists, such that the sum of the digits of n raised to consecutive powers starting from p is equal to k * n.

In other words, writing the consecutive digits of n as a, b, c, d ..., is there an integer k such that :

If it is the case we will return k, if not return -1.

Note: n and p will always be strictly positive integers.

Examples:

n = 89; p = 1 ---> 1 since 8¹ + 9² = 89 = 89 * 1

n = 92; p = 1 ---> -1 since there is no k such that 9¹ + 2² equals 92 * k

n = 695; p = 2 ---> 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2

n = 46288; p = 3 ---> 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51