22055_StuPId

At DUT, the Dreamland University of Technology, all students have personal id, numbers with six or seven digits. But they're not just any numbers. Only those that have a checksum with a zero as last digit can be valid ids.
Problem
Here's how to compute the checksum of an id number. Multiply the digits from back to front (!) with repeating factors 9, 3, 7. Then simply add the products. Example:
id number :  1  3  9  0  2  7  2
factors   :  9  7  3  9  7  3  9
products  :  9 21 27  0 14 21 18

Here the checksum is 9+21+27+0+14+21+18 = 110. The last digit is zero, so the id is valid. Sometimes students have very bad handwriting and the teaching assistents have a hard time identifying the id’s. You're asked to help in special cases, where exactly one digit is unreadable. In that case, the missing digit can be computed (there's always exactly one correct digit, thanks to 9, 3 and 7 being relatively prime to 10). Note that the students always begin concentrated and thus the first digit will always be readable (and not zero).

4
13?0272
3?5678
345?78
314?592

Scenario #1:
1390272

Scenario #2:
335678

Scenario #3:
345778

Scenario #4:
3146592

Huge input and output, scanf and printf are recommended.