dive it out

how many numbers greater than 1 exist such that the number is equal to the sum of its digits raised to the power the adjacent left digit of the former (starting from right) . the power which the left most digit is to raised is the right most digit. thus completing the loop for example lets take a 4 digit number abcd where a,b,c,d represents the digits of the number .

now according to the above statement find the number abcd such that abcd= (d^c)+(c^b)+(b^a)+(a^d) .

( this given example is set for four digit number but you have to find if any n digit number exist obeying the given rule, n is a natural number)