CF1305C.Kuroni and Impossible Calculation

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 $\prod_{1\le i \lt j\le n} |a_i - a_j|$. As result can be very big, output it modulo $m$.

If you are not familiar with short notation, $\prod_{1\le i \lt j\le n} |a_i - a_j|$ 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 $1\le i \lt j \le n$.

要成为 Codeforces 的王者，Kuroni 必须解决以下问题。

他被给定 $n$ 个数 $a_1, a_2, \dots, a_n$。请帮助 Kuroni 计算 $\prod_{1\le i \lt j\le n} |a_i - a_j|$。由于结果可能非常大，请输出其对 $m$ 取模的值。

若你不熟悉这种简写记号，则 $\prod_{1\le i \lt j\le n} |a_i - a_j|$ 等于 $|a_1 - a_2|\cdot|a_1 - a_3|\cdot$ $\dots$ $\cdot|a_1 - a_n|\cdot|a_2 - a_3|\cdot|a_2 - a_4|\cdot$ $\dots$ $\cdot|a_2 - a_n| \cdot$ $\dots$ $\cdot |a_{n-1} - a_n|$。换言之，这是对所有满足 $1\le i \lt j \le n$ 的 $|a_i - a_j|$ 的乘积。

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 $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$).

第一行包含两个整数 $n$、$m$（$2\le n \le 2\cdot 10^5$，$1\le m \le 1000$）——分别为数字的个数和模数。

第二行包含 $n$ 个整数 $a_1, a_2, \dots, a_n$（$0 \le a_i \le 10^9$）。

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

输出单个数字 — $\prod_{1\le i \lt j\le n} |a_i - a_j| \bmod m$。

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

在第一个样例中，$|8 - 5| = 3 \equiv 3 \bmod 10$。

在第二个样例中，$|1 - 4|\cdot|1 - 5|\cdot|4 - 5| = 3\cdot 4 \cdot 1 = 12 \equiv 0 \bmod 12$。

在第三个样例中，$|1 - 4|\cdot|1 - 9|\cdot|4 - 9| = 3 \cdot 8 \cdot 5 = 120 \equiv 1 \bmod 7$。