1305C · Kuroni and Impossible Calculation

Problem: To become the king of Codeforces, Kuroni has to solve the following problem.

He is given $n$ numbers $a_1, a_2, \dots, a_n$. Help Kuroni to calculate $\prod_{1\le i<j\le n} |a_i - a_j|$. As result can be very big, output it modulo $m$.

If you are not familiar with short notation, $\prod_{1\le i<j\le n} |a_i - a_j|$ is equal to $|a_1 - a_2|\cdot|a_1 - a_3|\cdot$ $\dots$ $\cdot|a_1 - a_n|\cdot|a_2 - a_3|\cdot|a_2 - a_4|\cdot$ $\dots$ $\cdot|a_2 - a_n| \cdot$ $\dots$ $\cdot |a_{n-1} - a_n|$. In other words, this is the product of $|a_i - a_j|$ for all $1\le i < j \le n$.

Input Format: The first line contains two integers $n$, $m$ ($2\le n \le 2\cdot 10^5$, $1\le m \le 1000$) — number of numbers and modulo.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$).

Output Format: Output the single number — $\prod_{1\le i<j\le n} |a_i - a_j| \bmod m$.

Note: In the first sample, $|8 - 5| = 3 \equiv 3 \bmod 10$.

In the second sample, $|1 - 4|\cdot|1 - 5|\cdot|4 - 5| = 3\cdot 4 \cdot 1 = 12 \equiv 0 \bmod 12$.

In the third sample, $|1 - 4|\cdot|1 - 9|\cdot|4 - 9| = 3 \cdot 8 \cdot 5 = 120 \equiv 1 \bmod 7$.