Bootstrap / startup switch circuit for 3.3V from 9V ?

[smoothVTer]

My system has a 9V battery as the main power supply.   There is a 4-position switch as well;  in the system off state, position 0,  the switch poles are open.    Positions 1, 2, and 3 short the switch's poles to ground individually.    This allows me to interface the switch with GPIOs on my uC for function selection, by virtue of external pull-up resistors on the uC  GPIO pins. 

The LDO has an enable pin which must be driven logic high. I want this enable pin driven high when the switch pulls any of the GPIO lines to ground.   The problem is, I cannot think of a circuit which will allow me to use the 9V (always-on) supply and interface with the 3.3V side of the uC without drawing current into the uC while in shutdown.

Here's an illustration.  Only a single switch pole is shown, but the idea is the same between all the GPIO lines.  I've modeled the 3 GPIOs as typical ESD rail to rail diodes & 50p / 1meg input impedance:



What circuitry can I use to achieve this function?   In the off state, the VDD supply is sitting at 0V, so I cannot rely on VDD to power any logic gates.  I could power logic gates from the 9V supply, but then I have 2 problems:   a level mismatch between the GPIO expectation of 3.3V nominal logic levels, and the 3.3V and 9V supplies fighting for dominance.    There must be a way to bridge the 9V supply leg with the GPIO lines, perhaps by sensing current from the 9V rail to ground, but I cannot come up with a circuit to do this safely for the GPIO lines.