Problem:
Score : 800 points

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.

Constraints

2 \leq N \leq 200\,000

0 \leq A_i < P = 200\,003

All values in input are integers.

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).

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

Solution in Pascal (FPC 3.0.4) (judge verdict: Time Limit Exceeded)

program Hello;

var n,i,j:longint;
    t : int64;
    a :array[1..200001] of longint;
begin
   readln(n);
   for i:=1 to n-1 do
    begin
       read(a[i]);
    end;
   readln(a[n]);
   for i:=1 to n-1 do
    for j:=i+1 to n do
     t:=t+((a[i]*a[j]) mod 200003);
    write(t);
    readln;
end.