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;

struct comp {
  double re, im;

  comp(double x = 0, double y = 0) : re(x), im(y) {}

  comp operator+(const comp& a) { return comp(re + a.re, im + a.im); }
  comp operator-(const comp& a) { return comp(re - a.re, im - a.im); }
  comp operator*(const comp& a) { return comp(re * a.re - im * a.im, re * a.im + im * a.re); }
};

const double pi = acos(-1);

void fft(vector<comp>& x) {
  int n = x.size();
  for (int i = 1, k = 0; i < n; ++i) {
    for (int bit = n >> 1; (k ^= bit) < bit; bit >>= 1) {}
    if (i < k) {
      swap(x[i], x[k]);
    }
  }
  for (int k = 1; k < n; k <<= 1) {
    comp wlen(cos(pi / k), sin(pi / k));
    for (int i = 0; i < n; i += (k << 1)) {
      comp w(1);
      for (int j = 0; j < k; ++j) {
        comp tmp = w * x[i + j + k];
        x[i + j + k] = x[i + j] - tmp;
        x[i + j] = x[i + j] + tmp;
        w = w * wlen;
      }
    }
  }
}

template<class T> vector<T> mul(vector<T>& a, vector<T>& b) {
  int n = max(a.size(), b.size());
  int lg = 31 - __builtin_clz(n);
  n = 1 << (lg + 1 + (n != (1 << lg)));
  vector<comp> fa(n), fb(n);
  for (int i = 0; i < a.size(); ++i) {
    fa[i] = a[i];
  } for (int i = 0; i < b.size(); ++i) {
    fb[i] = b[i];
  }
  fft(fa);
  if (a == b) {
    fb = fa;
  } else {
    fft(fb);
  }
  for (int i = 0; i < n; ++i) {
    fa[i] = fa[i] * fb[i];
  }
  fft(fa);
  reverse(fa.begin() + 1, fa.end());
  vector<T> c(n);
  for (int i = 0; i < n; ++i) {
    c[i] = (T)(fa[i].re / n + (fa[i].re > 0 ? 0.5 : -0.5));
  }
  return c;
}

const int P = 2e5 + 3;
int p[P];
int ord[P];

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  p[0] = 1;
  for (int i = 1; i < P; ++i) p[i] = p[i - 1] * 2 % P;
  for (int i = 0; i < P - 1; ++i) ord[p[i]] = i;
  //for (int i = 1; i < 10; ++i) cout << ord[i] << endl;
  vector<long long> cnt(P);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i]) ++cnt[ord[a[i]]];
  }
  cnt = mul(cnt, cnt);
  long long ans = 0;
  for (int i = 0; i < P - 1; ++i) ans += 1ll * p[i] * (cnt[i] + cnt[i + P - 1]);
  for (int i = 0; i < n; ++i) {
    ans -= 1ll * a[i] * a[i] % P;
  }
  cout << ans / 2;

  return 0;
}