# deterministic version takes 0.5.max(0, n-1).max(0, n-2)
# probabilistic version takes (1/p).0.5.max(0, n-1).max(0, n-2)

def f():
    var i, j, n

    i = 1
    while i < n:
        j = i - 1
        while j > 0:
            prob(1,2):
                j = j - 1
            tick 1
        i = i + 1