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 Go (1.14.1) (judge verdict: Time Limit Exceeded)

package main

import (
	"fmt"
)

func main() {
	var n,d int

	P:=200003

	fmt.Scanf("%d", &n)
	slice := make([]int,n,n)
	for i := 0; i < n; i++ {
		fmt.Scanf("%d", &d)
		slice[i] = d
	}
	total:=0
	tmp:=0
	for i:=0;i<n-1;i++ {
		for j:=1+i;j<n;j++ {
			tmp=slice[i]*slice[j]
			if (tmp<P) {
				total += tmp
			} else {
				total += tmp%P
			}
			
		}
	}
	fmt.Println(total)
}