Koshary

Observation 1

Long steps are boring in the best possible way: they add exactly $2$ to one coordinate.

So from $(0,0)$, after any number of long steps, Yousef can reach any point of the form

$$(2a, 2b)$$

for nonnegative integers $a,b$.

In other words, with only long steps, both coordinates must stay even. Parity does not budge. Zero drama.

Observation 2

A short step adds $1$ to exactly one coordinate, and Yousef may use it at most once.

That means during the whole journey, he can flip the parity of at most one coordinate.

So the final reachable parity patterns are exactly:

$$(\text{even},\text{even}),\quad (\text{odd},\text{even}),\quad (\text{even},\text{odd}).$$

The one pattern that cannot happen is

$$(\text{odd},\text{odd}),$$

because that would require flipping both coordinates’ parity, and he only gets one short step. Life is cruel.

Conclusion

Yousef can reach $(x,y)$ iff not both $x$ and $y$ are odd.

So the condition is simply:

$$\text{reachable} \iff \neg(x\text{ is odd }\land y\text{ is odd}).$$

Equivalently:

• if $x$ and $y$ are both odd, print NO
• otherwise, print YES

Why this is sufficient

If both are even, just use long steps only.

If exactly one is odd, say $x$ is odd and $y$ is even, then:

• use long steps to reach $(x-1, y)$, which has even coordinates,
• then use one short step in the $x$ direction.

Same idea if $y$ is the odd one.

So the parity condition is not just necessary — it’s also sufficient.

Complexity

Each test case is answered in constant time:

$$O(1)$$

For $t$ test cases, total complexity is:

$$O(t)$$

with

$$O(1)$$

extra memory.

Nice little implementation/math problem. Blink and you’ll overcomplicate it.