Problem Statement

Let’s take a prime P = 200\,003.
You are given $N integers A_1, A_2, \ldots, A_N$.
Find the sum of $((A_i \cdot A_j) \bmod P)$ over all $N \cdot (N-1) / 2$ unordered pairs of elements (i < j).
Please note that the sum isn't computed modulo P.

Input

Input is given from Standard Input in the following format.
N
$A_1 A_2 \cdots A_N$

Output

Print one integer — the sum over $((A_i \cdot A_j) \bmod P)$.

Constraints

• $2 \leq N \leq 200$\,000
• $0 \leq A_i < P = 200$\,003
• All values in input are integers.

Sample Input 1

4
2019 0 2020 200002

Sample Output 1

474287

The non-zero products are:

• $2019 \cdot 2020 \bmod P = 78320$
• $2019 \cdot 200002 \bmod P = 197984$
• $2020 \cdot 200002 \bmod P = 197983$

So the answer is 0 + 78320 + 197984 + 0 + 0 + 197983 = 474287.

Sample Input 2

5
1 1 2 2 100000

Sample Output 2

600013

见题面 Input 部分

见题面 Output 部分