Recursion-Last Index Of a Number in an Array

Given an array of length N and an integer x, you need to find and return the last index of integer x present in the array. Return -1 if it is not present in the array.
Last index means - if x is present multiple times in the array, return the index at which x comes last in the array.
You should start traversing your array from 0, not from (N - 1).
Do this recursively. Indexing in the array starts from 0.
Input Format :
Line 1 : An Integer N i.e. size of array
Line 2 : N integers which are elements of the array, separated by spaces
Line 3 : Integer x
Output Format :
last index or -1
Constraints :
1 <= N <= 10^3
Sample Input :
4
9 8 10 8
8
Sample Output :
3

****************************Code*******************************

public class Solution {

	public static int lastIndex(int input[], int x) {
  
		int li=input.length-1;
		return lastIndex(input,x,li);
	}
	private static int lastIndex(int input[],int x,int li)
	{
		if(li<0)
			return -1;
		if(input[li]==x)
			return li;
		int ans=lastIndex(input,x,li-1);
		return ans;
	}
	
}