Problem7358--ABC275 —— G - Infinite Knapsack

7358: ABC275 —— G - Infinite Knapsack

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Time Limit : 1.000 sec  Memory Limit : 256 MiB

Description

There are N kinds of items, each with infinitely many copies. The i-th kind of item has a weight of $A_i$, a volume of $B_i$, and a value of $C_i$.
Level X Takahashi can carry items whose total weight is at most X and whose total volume is at most X. Under this condition, it is allowed to carry any number of items of the same kind, or omit some kinds of items.
Let f(X) be the maximum total value of items Level X Takahashi can carry. It can be proved that the limit $\displaystyle\lim_{X\to \infty} \frac{f(X)}{X}$ exists. Find this limit.

Input

The input is given from Standard Input in the following format:
$N$
$A_1\ B_1\ C_1$
$A_2\ B_2\ C_2$
$\vdots$
$A_N\ B_N\ C_N$

Output

Print the answer. Your output will be considered correct if the absolute or relative error from the judge's answer is at most $10^{-6}$.

Constraints

$1≤N≤2×10^5$
$10^8\leq A_i,B_i,C_i \leq 10^9$
All values in the input are integers.

Sample 1 Input

2
100000000 200000000 100000000
200000000 100000000 100000000

Sample 1 Output

0.6666666666666667

When X=300000000, Takahashi can carry items whose total weight is at most 300000000 and whose total volume is at most 300000000.

He can carry, for instance, one copy of the 1-st item and one copy of the 2-nd item. Then, the total value of the items is 100000000+100000000=200000000. This is the maximum achievable value, so $\dfrac{f(300000000)}{300000000}=\dfrac{2}{3}$.

It can also be proved that $\displaystyle\lim_{X\to \infty} \frac{f(X)}{X}$ equals $\dfrac{2}{3}$. Thus, the answer is 0.6666666...

Sample 2 Input

1
500000000 300000000 123456789

Sample 2 Output

0.2469135780000000

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