For every positive integer n, let mod5(n) be the remainder obtained when n is divided by 5. Define a function f:{0,1,2,3,…}×{0,1,2,3,4}→{0,1,2,3,4} recursively as follows:
f(i,j)=⎩⎨⎧mod5(j+1)f(i−1,1)f(i−1,f(i,j−1))if i=0 and 0≤j≤4if i≥1 and j=0, andif i≥1 and 1≤j≤4
What is f(2015,2)?<spanclass=′latex−bold′>(A)</span>0<spanclass=′latex−bold′>(B)</span>1<spanclass=′latex−bold′>(C)</span>2<spanclass=′latex−bold′>(D)</span>3<spanclass=′latex−bold′>(E)</span>4