6 3 5
Yes
2 4 5 3
When choosing $P = (2, 4, 5, 3)$, perform four operations on $A = (0, 1, 2, 3, 4, 5)$ as follows.
* The remainder when $P_{0} = 2$ is divided by $|A| = 6$ is $2$, so remove $A_{2} = 2$ from $A$. Then, $A = (A_{0}, A_{1}, A_{2}, A_{3}, A_{4}) = (0, 1, 3, 4, 5)$.
* The remainder when $P_{1} = 4$ is divided by $|A| = 5$ is $4$, so remove $A_{4} = 5$ from $A$. Then, $A = (A_{0}, A_{1}, A_{2}, A_{3}) = (0, 1, 3, 4)$.
* The remainder when $P_{2} = 5$ is divided by $|A| = 4$ is $1$, so remove $A_{1} = 1$ from $A$. Then, $A = (A_{0}, A_{1}, A_{2}) = (0, 3, 4)$.
* The remainder when $P_{3} = 3$ is divided by $|A| = 3$ is $0$, so remove $A_{0} = 0$ from $A$. Then, $A = (A_{0}, A_{1}) = (3, 4)$.
Therefore, the goal can be achieved by setting $P = (2, 4, 5, 3)$. It is also acceptable to output $P = (5, 2, 4, 3)$.9 2 4
Yes 4 5 6 7 3 8 2
178 68 167
No
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