One can get accepted by embedding the information of \(15 \times 15\) grid in the source code, but it’s a bit troublesome. Since the grid is symmetric and regular with respect to the center square (at the \(8\)-th row and \(8\)-th column), we can solve it more simply as follows.

In the given grid, the square is painted black if the “distance” from the central square is odd; white if it is even. Specifically, the square at the \(R\)-th row and \(C\)-th column is black if and only if its Chebyshev distance (chessboard distance) from the central square,

\[\max\lbrace |R-8|, |C-8|\rbrace, \tag{1}\]

is odd. (Here, \(|x|\) denotes the absolute value of \(x\).) Therefore, the problem can be concisely solved by computing (1) above and checking its parity, as in the following sample code (C++):

#include <iostream>
using namespace std;

int main(void)
{
  int r, c;
  cin >> r >> c;
  
  if(max(abs(r-8), abs(c-8)) % 2) cout << "black" << endl;
  else cout << "white" << endl;
  
  return 0;
}