N Queens using Bit manipulation

With Bit manipulation the N Queens problem runs quite faster than the trivial method of solving N Queens where we always have to check whether the queen can be placed in the current cell (defined as place(row, col) say), & this checking is also not made in constant time! But using Bit manipulation we are only traversing through the cells where a queen can be placed!! i.e. we are not only removing the cost for place(row, col) but also we are completely ignoring the cells where a queen cannot be placed.

I ran N Queens for N = 14 and it took less than 1s to run. While trivial method of cost O(N!) took > 15s

import java.util.Arrays;

class NQueens {
    static int N, count;
    static int[] column;
    
    public static void main (String[] args) {
        N = 5;
        column = new int[N];
        nQueens(0, 0, 0, 0);
        System.out.println(count);
    }
    
    private static void nQueens(int row, int ltDiag, int rtDiag, int down) {
        if (row == N) {
            System.out.println(Arrays.toString(column));
            count++;
            return;
        }
        int mask = ltDiag | rtDiag | down;
        while (mask < (1 << N) - 1) {
            int col = ~mask & mask + 1; // gets rightmost 0-bit
            column[row] = Integer.numberOfTrailingZeros(col);
            int lD = (ltDiag | col) >> 1;
            int rD = (rtDiag | col) << 1 & (1 << N) - 1;
            nQueens(row + 1, lD, rD, down | col);
            mask |= col;
        }
    }
}

Output

[0, 2, 4, 1, 3]
[0, 3, 1, 4, 2]
[1, 3, 0, 2, 4]
[1, 4, 2, 0, 3]
[2, 0, 3, 1, 4]
[2, 4, 1, 3, 0]
[3, 0, 2, 4, 1]
[3, 1, 4, 2, 0]
[4, 1, 3, 0, 2]
[4, 2, 0, 3, 1]
10

The column[] contains the solution, where column[i] gives where in the i^th row the Queen will be placed (Zero based indexing)