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<bits/stdc++.h>

using namespace std;

typedef pair<int, int> ii;
typedef long long ll;

#define int ll

#define REP(i, n) for(int i = 0; i < (n); i++)
typedef int llint;
namespace FFT {
	const int MAX = 1 << 19;

	typedef llint value;
	typedef complex<double> comp;

	int N;
	comp omega[MAX];
	comp a1[MAX], a2[MAX];
	comp z1[MAX], z2[MAX];

	void fft(comp *a, comp *z, int m = N) {
		if (m == 1) {
			z[0] = a[0];
		} else {
			int s = N/m;
			m /= 2;

			fft(a, z, m);
			fft(a+s, z+m, m);

			REP(i, m) {
				comp c = omega[s*i] * z[m+i];
				z[m+i] = z[i] - c;
				z[i] += c;
			}
		}
	}

	void mult(value *a, value *b, value *c, int len) {
		N = 2*len;
		while (N & (N-1)) ++N;
		assert(N <= MAX);

		REP(i, N) a1[i] = 0;
		REP(i, N) a2[i] = 0;
		REP(i, len) a1[i] = a[i];
		REP(i, len) a2[i] = b[i];

		REP(i, N) omega[i] = polar(1.0, 2*M_PI/N*i);
		fft(a1, z1, N);
		fft(a2, z2, N);

		REP(i, N) omega[i] = comp(1, 0) / omega[i];
		REP(i, N) a1[i] = z1[i] * z2[i] / comp(N, 0);
		fft(a1, z1, N);

		REP(i, 2*len) c[i] = round(z1[i].real());
	}

	void mult_mod(int *a, int *b, int *c, int len, int mod) {
		static llint a0[MAX], a1[MAX];
		static llint b0[MAX], b1[MAX];
		static llint c0[MAX], c1[MAX], c2[MAX];

		REP(i, len) a0[i] = a[i] & 0xFFFF;
		REP(i, len) a1[i] = a[i] >> 16;

		REP(i, len) b0[i] = b[i] & 0xFFFF;
		REP(i, len) b1[i] = b[i] >> 16;

		FFT::mult(a0, b0, c0, len);
		FFT::mult(a1, b1, c2, len);

		REP(i, len) a0[i] += a1[i];
		REP(i, len) b0[i] += b1[i];
		FFT::mult(a0, b0, c1, len);
		REP(i, 2*len) c1[i] -= c0[i] + c2[i];

		REP(i, 2*len) c1[i] %= mod;
		REP(i, 2*len) c2[i] %= mod;
		REP(i, 2*len) c[i] = (c0[i] + ((long long) c1[i] << 16) + ((long long) c2[i] << 32)) % mod;
	}
}
#undef REP


const int MOD = 2e5+3, G = 2;

int inv_pw[MOD], pw[MOD], c[MOD], mul[2*MOD];

signed main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
	int it=1, p=0, ans=0;
	do{
		inv_pw[it] = p ++;
		it = it * G % MOD;
	}while(it != 1);
	pw[0] = 1;
	for(int i=1 ; i<MOD ; i++)
		pw[i] = pw[i-1] * G % MOD;
	int n;cin >> n;
	for(int i=0 ; i<n ; i++){
		int d;cin >> d;
		ans -= d * d % MOD;
		if(d == 0)
			continue;
		c[ inv_pw[d] ] ++;
	}
	FFT::mult(c, c, mul, MOD);
	for(int i=0 ; i<2*MOD ; i++){
		int cur = i % (MOD-1);
		ans += mul[i] * pw[cur];
	}
	cout << ans/2 << "\n";
    return 0;
}