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 C (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void){
    int P = 200003;
    int n = 0;
    int a[200000];
    int sum = 0;
    int prdct = 0;

    scanf("%d",&n);

    for(int i = 0; i< n; i++){
        scanf("%d",&a[i]);
    }

    for(int i = 0; i < n; i++){
        if(a[i] != 0){
            for(int j = i + 1; j < n; j++){
                if(a[j] != 0){
                    prdct = a[i] * a[j];
                    sum += prdct % P;
                }
            }
        }
    }

    printf("%d", sum);
    return 0;
}