Four Vertices of a Cube

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Time limit1sMemory limit1024 MB

Summary

Given T test cases of four distinct indices of a unit cube's vertices, determine whether each set of four points lies on a single face of the cube.

Level: Easy2 of 10
Topics: Implementation, Brute force
Solved: No attempts yet

Problem

Eight points in three-dimensional coordinate space, $P\_0 = (0,0,0)$ , $P\_1 = (0,0,1)$ , $P\_2 = (0,1,0)$ , $P\_3 = (0,1,1)$ , $P\_4 = (1,0,0)$ , $P\_5 = (1,0,1)$ , $P\_6 = (1,1,0)$ , and $P\_7 = (1,1,1)$ , are the vertices of a cube $S$ .

Determine whether four given vertices lie on one face of the cube $S$ .

Input

The first line contains the number of test cases $T$ . $( 1 \le T \le 2 000 )$

Each of the following $T$ lines contains four distinct integers $a$ , $b$ , $c$ , and $d$ , separated by spaces. Each integer is the index of a vertex of the cube $S$ . $(0 \le a, b, c, d \lt 8)$

Output

For each test case, print one line. Print YES if $P\_a$ , $P\_b$ , $P\_c$ , and $P\_d$ are four vertices on one face of the cube $S$ . Otherwise, print NO.

Examples1

Example 1
Input
```
2
5 0 4 1
1 2 6 5
```
Expected output
```
YES
NO
```