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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

long long P = 200003;

int main() {
	int n;
	cin >> n;
	long long ans = 0;
	
	vector<long long> a(n);
	
	for(long long& el : a) {
		cin >> el;
		el %= P;
	}
	
	for(int i = 0; i < n; i++) {
		for(int j = i+1; j < n; j++) {
			ans += a[i]*a[j]%P;
		} 
	}
	
	cout << ans;
}