There is an operation program, when a ⊕ B = n (n is a constant), define (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known 1 ⊕ 1 = 2, then 2010 ⊕ 2010=______ ．

There is an operation program, when a ⊕ B = n (n is a constant), define (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known 1 ⊕ 1 = 2, then 2010 ⊕ 2010=______ ．

From a ⊕ B = n, (a + 1) ⊕ B = n + 1, a ⊕ B (B + 1) = n-2, and 1 ⊕ 1 = 2, we get 2 ⊕ 1 = 2, 2 ⊕ B = n, (a + 1) ⊕ B = n + 1, (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, and 1 ⊕ (B + 1) 1 = 1 = 1 = 2 = 2, and 1 ⊕ 1 = 2 = 2-2 = 2 = 2 = 3, 2,4 ⊕3 = 1 = 1-1, 1,4 ⊕4 = 4 = 1,4 = 1-4 = 1-2-2-+ 1 = - 1, 6 ⊕ 6 = - 1-2 = - 3 ∴2010⊕2010=-200...

There is an operation program, when a ⊕ B = n (n is a constant), define (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known 1 ⊕ 1 = 2, then 2010 ⊕ 2010=______ ．

From a ⊕ B = n, (a + 1) ⊕ B = n + 1, a ⊕ B (B + 1) = n-2, and 1 ⊕ 1 = 2, we get 2 ⊕ 1 = 2, 2 ⊕ B = n, (a + 1) ⊕ B = n + 1, (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, and 1 ⊕ (B + 1) 1 = 1 = 1 = 2 = 2, and 1 ⊕ 1 = 2 = 2-2 = 2 = 2 = 3, 2,4 ⊕3 = 1 = 1-1, 1,4 ⊕4 = 4 = 1,4 = 1-4 = 1-2-2-+ 1 = - 1, 6 ⊕ 6 = - 1-2 = - 3 ∴2010⊕2010=-200...