Possible?

1    12    13    14    15    16    17    18    19    110    111    112=0\large 1 \; \square \; \dfrac{1}{2} \; \square \; \dfrac{1}{3}\; \square \; \dfrac{1}{4}\; \square \; \dfrac{1}{5} \; \square \; \dfrac{1}{6} \; \square \; \dfrac{1}{7} \; \square \; \dfrac{1}{8} \; \square \; \dfrac{1}{9} \; \square \; \dfrac{1}{10} \; \square \; \dfrac{1}{11} \; \square \; \dfrac{1}{12}= 0

Can you fill the boxes with + and + \text{ and } - to make this equation true?

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