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Look at the most inner problem,
1 + 2 0 1 1 ∗ 2 0 1 3 = 1 + ( 2 0 1 2 − 1 ) ( 2 0 1 2 + 1 )
easily, we could simplify it to 2 0 1 2 2 .
By doing this method three times, we will have 2 0 0 9 2 = 2 0 0 9 .