Optimize what registered cron jobs cost

A cron job that dispatches to the model pays for its whole injected context on every wake, whether or not that wake had anything to do. Two cheaper modes exist and neither reaches a job that was already registered without them.

Mode	What it costs	What it gives up
script or command job	no model turn at all	all reasoning, and the carried previous result
agent job with minimal_context	a model turn on a small context	memory, lessons, steering, skills, prior session history
plain agent job	a model turn on the full context	nothing

Your job in this skill is to move each of the user's jobs to the cheapest mode it can safely use, and to leave alone the ones that genuinely reason.

Run the scan first

The scan is deterministic, so it costs no inference. Run it before forming any opinion.

It reads nothing from disk. Two steps, and the first one matters:

1. Call cron_list with json set to true. That is the only reader that can see what the scan needs. The job store is scoped to the jobs your session owns, and the run history is readable only by the gateway, so a script cannot fetch either itself.
2. Write that JSON to a file and pass it in. Do NOT put the document on the command line -- a prompt inside it containing backticks or $(...) would be executed by the shell.

python3 <this skill's dir>/scripts/cron_cost_scan.py --input <the file you wrote>

Resolve <this skill's dir> from wherever you loaded this SKILL.md, rather than hardcoding a path. The data home moves with KIROCREW_HOME, so a hardcoded ~/.kiro/crew/skills/... is wrong for some users.

Useful flags:

• --json for machine-readable output when you need to fold the result into a table.
• --min-runs N to change how many recorded runs count as evidence. Default is 5.

To look at one job, narrow the cron_list call with its ids instead of filtering afterwards.