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;
const int mod = 200003;
int val[mod];

typedef long long LL;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

int get(int s) {
    return s > 1 ? 32 - __builtin_clz(s - 1) : 0;
}

typedef complex<double> base;

void fft(vector<base> &a, bool inv){
	int n = a.size(), j = 0;
	vector<base> roots(n/2);
	for(int i=1; i<n; i++){
		int bit = (n >> 1);
		while(j >= bit){
			j -= bit;
			bit >>= 1;
		}
		j += bit;
		if(i < j) swap(a[i], a[j]);
	}
	double ang = 2 * acos(-1) / n * (inv ? -1 : 1);
	for(int i=0; i<n/2; i++){
		roots[i] = base(cos(ang * i), sin(ang * i));
	}

	for(int i=2; i<=n; i<<=1){
		int step = n / i;
		for(int j=0; j<n; j+=i){
			for(int k=0; k<i/2; k++){
				base u = a[j+k], v = a[j+k+i/2] * roots[step * k];
				a[j+k] = u+v;
				a[j+k+i/2] = u-v;
			}
		}
	}
	if(inv) for(int i=0; i<n; i++) a[i] /= n;
}

vector<ll> DOIT(vector<ll> &v, vector<ll> &w){
	int n = 1;
	while(n < max(v.size(), w.size())) n <<= 1;
	n <<= 1;
	vector<base> v1(n), v2(n);
	vector<base> r1(n), r2(n);
	for(int i=0; i<v.size(); i++){
		v1[i] = base(v[i] >> 15, v[i] & 32767);
	}
	for(int i=0; i<w.size(); i++){
		v2[i] = base(w[i] >> 15, w[i] & 32767);
	}
	fft(v1, 0);
	fft(v2, 0);
	for(int i=0; i<n; i++){
		int j = (i ? (n - i) : i);
		base ans1 = (v1[i] + conj(v1[j])) * base(0.5, 0);
		base ans2 = (v1[i] - conj(v1[j])) * base(0, -0.5);
		base ans3 = (v2[i] + conj(v2[j])) * base(0.5, 0);
		base ans4 = (v2[i] - conj(v2[j])) * base(0, -0.5);
		r1[i] = (ans1 * ans3) + (ans1 * ans4) * base(0, 1);
		r2[i] = (ans2 * ans3) + (ans2 * ans4) * base(0, 1);
	}
	fft(r1, 1);
	fft(r2, 1);
	vector<ll> ret(n);
	for(int i=0; i<n; i++){
		ll av = (ll)round(r1[i].real());
		ll bv = (ll)round(r1[i].imag()) + (ll)round(r2[i].real());
		ll cv = (ll)round(r2[i].imag());
		ret[i] = (av << 30) + (bv << 15) + cv;
	}
	while (ret.back() == 0) ret.pop_back();
	return ret;
}

int main() {
  int cur = 1;
  memset(val, -1, sizeof val);
  for(int i = 0; i < mod - 1; i++) {
    val[cur] = i;
    cur = (cur * 2) % mod;
  }
  int n;
  scanf("%d", &n);
  long long sum = 0;
  vector <ll> cnt (mod);
  for(int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if(x != 0) {
      cnt[val[x]] += 1;
    }
    sum -= (1LL * x * x) % mod;
  }
  auto ans = DOIT(cnt, cnt);

  cur = 1;
  for(int i = 0; i < ans.size(); i++) {
    sum += 1LL * cur * ans[i];
    cur = (cur * 2) % mod;
  }
  sum /= 2;
  printf("%lld\n", sum);
  return 0;
}