A magic square of order n is an arrangement of integers 1 through n2 in a square format such that the sum of numbers along any row, any column and any diagonal is the same. Your program inputs an odd integer n (< 50), and prints a magic square of order n. 

The following algorithm holds good only for an odd n. Assume n is an odd integer. 

Call the rectangular array magicSquare[][], and initialize all elements of the array to 0. Start by keeping 1 in the middle cell of row 0. That is, K = 1, I = 0, J = n/2 and magicSquare[I][J] = K. 

Once K is placed in a cell, do K++. Place the new K in the "north east" cell. In other words, after placing a K in a cell, do K++, I--, J++, and place the new K at the cell (I,J). 

There are four situations you cannot place the new K in the new cell: SituationÂ	SolutionÂ
The new (I,J) is out of bounds in the Northern direction i.e. I < 0.Â	Wrap around and go to the last row, but in the same column. i.e. I = n-1.Â
The new (I,J) is out of bounds in the Eastern direction. i.e. J > n-1.Â	Wrap around and go to column 0, but in the same row. i.e. J = 0.Â
The new (I,J) is out of bounds in both directions (happens only once). i.e. I < 0 and J> n-1.Â	Go one row below from the original position (before you did I-- and J++), in the same column. i.e. I +=2, J--.Â
The new (I,J) position in the array is already taken by an earlier K. i.e. magic_square[I][J] != 0.Â	Go one row below from the original position (before you did I-- and J++), in the same column. i.e. I +=2, J--.Â

Continue this process until all the numbers 1 - n2 (n squared) are placed