# PROBLEM LINK:

Practice

Contest: Division 1

Contest: Division 2

Contest: Division 3

Contest: Division 4

* Author:* irmuun

*iceknight1093*

**Tester & Editorialist:**# DIFFICULTY:

1182

# PREREQUISITES:

None

# PROBLEM:

Is it possible to tile a rectangular floor with length L and width W, using only rectangles whose perimeter *isn’t* divisible by 4?

# EXPLANATION:

Recall that the perimeter of a rectangle whose side lengths are a and b, is 2\cdot (a + b).

For this to be *not* divisible by 4, we’d need (a+b) to be odd — which means exactly one of a and b must be even, and the other must be odd.

In particular, this means the area of such a rectangle (which equals a\cdot b) must be even.

Now, let A = L\cdot W be the total area of the floor.

Suppose we tile it with k rectangles, and their areas are A_1, A_2, \ldots, A_k.

Of course, since it’s a tiling, A = A_1 + A_2 + \ldots + A_k must hold.

For the tiling to be valid, we need each A_i to be even, as discussed earlier.

Notice that this forces A to also be even, as the sum of some even numbers.

So, if the area A = L\cdot W is odd, the answer is immediately `No`

.

On the other hand, if A is even, the answer is always `Yes`

!

This is because a rectangle with even area can always be tiled using 1\times 2 (or 2\times 1) tiles, which have perimeter 6 and so are usable.

# TIME COMPLEXITY

\mathcal{O}(1) per test case.

# CODE:

## Editorialist's code (Python)

```
for _ in range(int(input())):
w, l = map(int, input().split())
print('Yes' if (w*l)%2 == 0 else 'No')
```