**Three Boxes Solution**

Chef has 3 boxes of sizes A, B, and C respectively. He puts the boxes in bags of size D (A≤B≤C≤D). Find the minimum number of bags Chef needs so that he can put each box in a bag. A bag can contain more than one box if the sum of sizes of boxes in the bag does not exceed the size of the bag.

**Input Format**

- The first line contains T denoting the number of test cases. Then the test cases follow.
- Each test case contains four integers A, B, C, and D on a single line denoting the sizes of the boxes and bags.

**Output Format**

For each test case, output on a single line the minimum number of bags Chef needs.

**Constraints**

- 1≤T≤100
- 1≤A≤B≤C≤D≤100

**Subtasks**

**Subtask 1 (100 points):** Original constraints

**Sample Input 1**

```
3
2 3 5 10
1 2 3 5
3 3 4 4
```

**Sample Output 1**

```
1
2
3
```

**Explanation**

**Test case 1**: The sum of sizes of boxes is 2+3+5=10 which is equal to the size of a bag. Hence Chef can put all three boxes in a single bag.**Test case 2**: Chef can put boxes of size 1 and 3 in one bag and box of size 2 in another bag.**Test case 3**: Chef puts all the boxes in separate bags as there is no way to put more than one box in a single bag.

**SOLUTION**

*Program Python:***Three Boxes Solution** in Python

```
for _ in range (int(input())):
a,b,c,d=map(int,input().split())
if a+b+c<=d:
print(1)
elif a+c<=d:
print(2)
elif b+a<=d:
print(2)
else:
print(3)
```

*October Long Challenge 2021*

*October Long Challenge 2021*