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 a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p 
- we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n. 
In other words:
> Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k
If it is the case we will return k, if not return -1.
**Note**: n and p will always be given as strictly positive integers.
```python
dig_pow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
dig_pow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
dig_pow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
dig_pow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
```
def dig_pow(n, p):
    digits = [int(d) for d in str(n)]
    
    sum = 0
    for i, d in enumerate(digits):
        sum += d ** (p+i)
    
    if sum % n == 0:
        return sum / n
    else:
        return -1
def dig_pow(n, p):
    digits = [int(d) for d in str(n)]
    
    sum = 0
    for i, d in enumerate(digits):
        sum += d ** (p+i)
    
    if sum % n == 0:
        return sum / n
    else:
        return -1