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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main(void){
  int a,i,j,sum;
  sum=0;
  int A[200000],b,c;
  scanf("%d",&a);
  for(i=0;i<a;i++){
    scanf("%d",&A[i]);
  }
  for(i=0;i<a-1;i++){
    for(j=i+1;j<a;j++){
      if((A[i]*A[j])<200003){
        sum+=A[i]*A[j];
        printf("%d",A[i]*A[j]);
      }
      else {b=(A[i]*A[j])%200003;
      
      sum+=b;}
    }
  }
  printf("%d",sum);
  return 0;
}