Wait the job ends before starts again

I'm looking for a method to schedule a job to run every 2 seconds but without to start another thread while the first one is running yet.
I read the FAQ and the issues but no one comment about it.

```
import threading
import time
import schedule

def job():
print("A: %s" % threading.get_ident(), time.time())
time.sleep(3)
print("B: %s" % threading.get_ident(), time.time())

def run_threaded(job_func):
job_thread = threading.Thread(target=job_func)
job_thread.start()

schedule.every(2).seconds.do(run_threaded, job)

while 1:
schedule.run_pending()
time.sleep(1)
```

The output is:
```
A: 139822667417344 1524702860
A: 139822659024640 1524702862
B: 139822667417344 1524702863
A: 139822667417344 1524702864
B: 139822659024640 1524702865
A: 139822659024640 1524702866
B: 139822667417344 1524702867
A: 139822667417344 1524702868
```

But, I would like something like it:
```
A: 139822667417344 1524702860
B: 139822667417344 1524702863
A: 139822667417344 1524702865
B: 139822667417344 1524702868
```

The `every(2)` start to count after the job ends.

Start by running the supplied example and reading the behavior of schedule.every(...).seconds.do and schedule.run_pending. Use the requested output sequence as the acceptance criterion, including the case where a job runs longer than its interval and must not overlap with the next run.

Written by the indexing model from the issue text.