Positive integers x and y satisfy 6x+7y=2012. Suppose the minimum value of x+y is p, and the maximum value is q, then p+q=how much.
正整数x，y满足6x+7y=2012．设x+y的最小值为p，最大值为q，则p+q=多少．
<gadget id="calculator">2_012 - 284</gadget>
<output>1_728</output>

<gadget id="calculator">1_728 / 6</gadget>
<output>288</output>

<gadget id="calculator">2_012 - 2</gadget>
<output>2_010</output>

<gadget id="calculator">2_010 / 6</gadget>
<output>335</output>

<gadget id="calculator">288 + 335</gadget>
<output>623</output>

<result>623</result>