Fast Food Restaurant
Contest: Codeforces Round 622 (Div. 2)
Difficulty rating: 900
Tags: brute force, greedy, implementation

Tired of boring office work, Denis decided to open a fast food restaurant.

On the first day he made $$$a$$$ portions of dumplings, $$$b$$$ portions of cranberry juice and $$$c$$$ pancakes with condensed milk.

The peculiarity of Denis's restaurant is the procedure of ordering food. For each visitor Denis himself chooses a set of dishes that this visitor will receive. When doing so, Denis is guided by the following rules:

- every visitor should receive at least one dish (dumplings, cranberry juice, pancakes with condensed milk are all considered to be dishes);
- each visitor should receive no more than one portion of dumplings, no more than one portion of cranberry juice and no more than one pancake with condensed milk;
- all visitors should receive different sets of dishes.

What is the maximum number of visitors Denis can feed?

Input format:
The first line contains an integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases to solve.

Each of the remaining $$$t$$$ lines contains integers $$$a$$$, $$$b$$$ and $$$c$$$ ($$$0 \leq a, b, c \leq 10$$$) — the number of portions of dumplings, the number of portions of cranberry juice and the number of condensed milk pancakes Denis made.

Output format:
For each test case print a single integer — the maximum number of visitors Denis can feed.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case of the example, Denis can feed the first visitor with dumplings, give the second a portion of cranberry juice, and give the third visitor a portion of cranberry juice and a pancake with a condensed milk.

In the second test case of the example, the restaurant Denis is not very promising: he can serve no customers.

In the third test case of the example, Denise can serve four visitors. The first guest will receive a full lunch of dumplings, a portion of cranberry juice and a pancake with condensed milk. The second visitor will get only dumplings. The third guest will receive a pancake with condensed milk, and the fourth guest will receive a pancake and a portion of dumplings. Please note that Denis hasn't used all of the prepared products, but is unable to serve more visitors.

Example input:
7
1 2 1
0 0 0
9 1 7
2 2 3
2 3 2
3 2 2
4 4 4

Example output:
3
0
4
5
5
5
7

Editorial:
Bruteforce solution
There are seven possible sets of dishes, so the simplest solution is to iterate over all possible $$$2^7$$$ subsets of sets of dishes.
You can also go over $$$7!$$$ permutations of sets of dishes and gather sets of dishes greedily in the selected order.
Greedy solution
Note that the solution can be optimal only when it is impossible to add an additional set of dishes to it.
Let the solution be such that it is impossible to add a single set of dishes to it and it does not have any set consisting of one dish, but there is a set consisting of two or three dishes containing this one dish. Then you can replace the corresponding set with a set of one dish, without worsening the answer. This means that at the beginning you can greedily add all the sets consisting of one dish.
The same can show that any set of three dishes can be replaced with a set of two dishes, so after the sets of one and two dishes are fixed, it is enough to simply check whether you can add a set of three dishes.
However, it is wrong to choose sets of two dishes greedily. Suppose that after choosing sets of one dish, there is one dish of the first type, one dish of the second type and two dishes of the third type. Then you can choose two sets of dishes, but if you take at the beginning a set of dishes of the first and second types, you won't get two different sets.
In this case, you can simply iterate over the order of choosing sets of two dishes or notice that all such tests have the form $$$2\, 2\, x$$$, $$$2\, x\, 2$$$, $$$x\, 2\, 2$$$, where $$$x \geq 3$$$, and solve them separately.