Problem: 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).

Examples

Input

4
2019 0 2020 200002


Output

474287


Input

5
1 1 2 2 100000


Output

600013

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <cmath>
#include <map>
using namespace std;
#define MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

const double PI = acos(-1);
struct CP {
	double x, y;
	CP(double x = 0, double y = 0) : x(x), y(y) {}
	CP operator+(const CP &a) { return CP(x + a.x, y + a.y); }
	CP operator-(const CP &a) { return CP(x - a.x, y - a.y); }
	CP operator*(const CP &a) { return CP(x * a.x - y * a.y, x * a.y + y * a.x); }
};
void FFTMakeG(int n, CP g[]) {
	for (int i = 0; i < n; ++i) { double th = PI * 2.0 * i / n; g[i] = CP(cos(th), sin(th)); }
}
void FFT(int n, CP g[], CP z[]) {
	int i, j, k, l, m;
	for (i = 1, j = 0; i < n; ++i) {
		for (k = n >> 1; (j ^= k) < k; k >>= 1);
		if (i < j) swap(z[i], z[j]);
	}
	for (l = k = 1, m = __builtin_ctz(n); m--; l |= k <<= 1) {
		for (i = 0; i < n; ++i) if (i < (j = i ^ k)) {
			CP t = z[j] * g[(i & l) << m];
			z[j] = z[i] - t;
			z[i] = z[i] + t;
		}
	}
}

const int P = 200003;
int N, A[202020];
i64 pows[202020];
map<i64, int> ord;
int cnt[202020];

CP g[1 << 19], xhi[1 << 19], xlo[1 << 19], y[1 << 19];
const int K = 330;
i64 ans[202020];

i64 naive()
{
	i64 ret = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = i + 1; j < N; ++j) ret += (A[i] * (i64)A[j]) % P;
	}
	return ret;
}

int main()
{
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) scanf("%d", A + i);

	i64 v = 1;
	for (int i = 0; i < P - 1; ++i) {
		ord[v] = i;
		pows[i] = v;
		v = v * 2 % P;
	}
	for (int i = 0; i < N; ++i) {
		int ai = A[i];
		if (ai == 0) continue;
		++cnt[ord[ai]];
	}
	FFTMakeG(1 << 19, g);
	for (int i = 0; i < P - 1; ++i) {
		xhi[i] = cnt[i] / K;
		xlo[i] = cnt[i] % K;
	}
	FFT(1 << 19, g, xhi);
	FFT(1 << 19, g, xlo);

	for (int i = 0; i < 3; ++i) {
		if (i == 0) {
			for (int j = 0; j < (1 << 19); ++j) y[j] = xhi[j] * xhi[j];
		} else if (i == 1) {
			for (int j = 0; j < (1 << 19); ++j) y[j] = xhi[j] * xlo[j];
		} else {
			for (int j = 0; j < (1 << 19); ++j) y[j] = xlo[j] * xlo[j];
		}
		reverse(g + 1, g + (1 << 19));
		FFT(1 << 19, g, y);
		reverse(g + 1, g + (1 << 19));

		i64 waffle = 1;
		if (i == 0) waffle = K * K;
		else if (i == 1) waffle = K * 2;

		for (int j = 0; j < (1 << 19); ++j) {
			i64 v = (i64)(y[j].x / (1 << 19) + 0.5);
			ans[j % (P - 1)] += v * waffle;
		}
	}
	i64 ret = 0;
	for (int i = 0; i < P - 1; ++i) {
		ret += ans[i] * pows[i];
		ret -= cnt[i] * pows[(i * 2) % (P - 1)];
	}
	ret /= 2;
	printf("%lld\n", ret);
	return 0;
}