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 long double ld;
typedef long long ll;

#define sz(x) (int)(x).size()
#define pb push_back
#define mp make_pair
#define f first
#define s second

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
typedef pair<int, int> pii;
typedef vector<int> vi;


template<typename T, typename U> bool ckmin(T &a, U b){ return b < a ? a = b, true : false; }
template<typename T, typename U> bool ckmax(T &a, U b){ return b > a ? a = b, true : false; }

const ll INF = 69696969;
const ll MOD = 2e5 + 3;

ll prod = 0;
int n;
//g[i] = 2^i, rg[i] = j s.t. 2^j = i
vector<double> poly, res;
vector<ll> g, rg;

typedef complex<double> C;
typedef vector<double> vd;
void fft(vector<C>& a) {
	int n = sz(a), L = 31 - __builtin_clz(n);
	static vector<complex<long double>> R(2, 1);
	static vector<C> rt(2, 1);  // (^ 10% faster if double)
	for (static int k = 2; k < n; k *= 2) {
		R.resize(n); rt.resize(n);
		auto x = polar(1.0L, acos(-1.0L) / k);
		rep(i,k,2*k) rt[i] = R[i] = i&1 ? R[i/2] * x : R[i/2];
	}
	vi rev(n);
	rep(i,0,n) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
	rep(i,0,n) if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int k = 1; k < n; k *= 2)
		for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
			// C z = rt[j+k] * a[i+j+k]; // (25% faster if hand-rolled)  /// include-line
			auto x = (double *)&rt[j+k], y = (double *)&a[i+j+k];        /// exclude-line
			C z(x[0]*y[0] - x[1]*y[1], x[0]*y[1] + x[1]*y[0]);           /// exclude-line
			a[i + j + k] = a[i + j] - z;
			a[i + j] += z;
		}
}
vd conv(const vd& a, const vd& b) {
	if (a.empty() || b.empty()) return {};
	vd res(sz(a) + sz(b) - 1);
	int L = 32 - __builtin_clz(sz(res)), n = 1 << L;
	vector<C> in(n), out(n);
	copy(all(a), begin(in));
	rep(i,0,sz(b)) in[i].imag(b[i]);
	fft(in);
	for (C& x : in) x *= x;
	rep(i,0,n) out[i] = in[-i & (n - 1)] - conj(in[i]);
	fft(out);
	rep(i,0,sz(res)) res[i] = imag(out[i]) / (4 * n);
	return res;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    n = 1;
    g.assign(MOD, INF);
    rg.assign(MOD, INF);
    for(int i = 0; i < MOD - 1; ++i){
        ckmin(g[i], n);
        ckmin(rg[n], i);
        n = (n * 2)%MOD;
    }
    cin >> n;
    poly.assign(MOD - 1, 0);
    for(int i = 0; i < n; ++i){
        ll x; cin >> x;
        if(!x) continue;
        // cout << rg[x] << "\n";
        poly[rg[x]]++;
        prod -= (x * x)%MOD;
    }
    res = conv(poly, poly);
    for(int i = 0; i < sz(res); ++i){
        prod += round(res[i]) * g[i%(MOD - 1)];
    }
    cout << (prod >> 1);
    return 0;
}