If log2(5+log3a)=3\log_{2}({5+\log_{3}{a}})=3log2(5+log3a)=3 and log5(4a+12+log2b)=3\log_{5}({4a+12+\log_{2}{b}})=3log5(4a+12+log2b)=3, then a + b is equal to
59
40
32
67
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