Consider the first test case, $$N = 3$$, $$ A = [3, 4, 1]$$. We want to evaluate $$f(1, 1) + f(1, 2) + f(2, 2) + f(2, 3) + f(3, 3)$$. Calculations are shown below:

$$f(1, 1)$$ : Replace $$A_1$$ with $$(2^{25} - 1)$$, we get $$A = [33554431, 4, 1]$$, Bitwise AND of elements is 0.
$$f(1, 2)$$ : Replace $$A_1, A_2$$ with $$(2^{25} - 1)$$, we get $$A = [33554431, 33554431, 1]$$, Bitwise AND of elements is 1.
$$f(2, 2)$$ : Replace $$A_2$$ with $$(2^{25} - 1)$$, we get $$A = [3, 33554431, 1]$$, Bitwise AND of elements is 1.
$$f(2, 3)$$ : Replace $$A_2, A_3$$ with $$(2^{25} - 1)$$, we get $$A = [3, 33554431, 33554431]$$, Bitwise AND of elements is 3.
$$f(3, 3)$$ : Replace $$A_3$$ with $$(2^{25} - 1)$$, we get $$A = [3, 4, 33554431]$$, Bitwise AND of elements is 0.
The sum is $$0 + 1 + 1 + 3 + 0 = 5$$.

Consider the second test case, $$N = 2$$, $$ A = [4, 2]$$. We want to evaluate $$f(1, 1) + f(2, 2) $$. Calculations are shown below:

$$f(1, 1)$$ : Replace $$A_1$$ with $$(2^{25} - 1)$$, we get $$A = [33554431, 2]$$, Bitwise AND of elements is 2.
$$f(2, 2)$$ : Replace $$A_2$$ with $$(2^{25} - 1)$$, we get $$A = [4, 33554431]$$, Bitwise AND of elements is 4.
The sum is $$2 + 4 = 6$$.