# see if we free invocations
resource f ()
	op p (int a) returns int  x

	int  j, n
	write (-9999)
	for [ i = 1 to 10000 ] {
		j = i
		n = p(i)
		if (n + j ~= 7) {
			write (i, j, n)
			stop
		}
	}
	write (-1111)


	proc p (a) returns x  {
		x = -a+7
		a = -876
	}
end
