#include <bits/stdc++.h>
#define sz(x) ((int)(x).size())
#define all(x) begin(x), end(x)
#ifdef memset0
#define log(...) fprintf(stderr, __VA_ARGS__)
#else
#define endl '\n'
#define log(...) (void(0))
#endif
using namespace std;
using ll = long long;
using lf = long double;
using ull = unsigned long long;

const int N = 200011;
vector<int> prime;
ll Qpow(ll a, ll p, ll mod) {
  ll res = 1;
  while (p) {
    if (p & 1) res = res * a % mod;
    a = a * a % mod, p >>= 1;
  }
  return res;
}
vector<int> a[N], now, mul;
string s;
bool isprime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      return false;
    }
  return true;
}
int main() {
#ifdef memset0
  freopen("A.in", "r", stdin);
#endif
  cin.tie(0)->sync_with_stdio(0);
  int p = 1.123e9;
  while (prime.size() < 5) {
    if (isprime(p)) {
      // cerr << p << endl;
      prime.push_back(p);
    }
    p++;
  }
  int n, l = prime.size();
  cin >> n;
  for (int k = 0; k < l; ++k) mul.emplace_back(1);
  for (int i = 1; i <= n; ++i) {
    cin >> s;
    now.clear();
    now.resize(l);
    for (int j = 0; j < s.size(); ++j)
      for (int k = 0; k < l; ++k) {
        now[k] = (now[k] * 10ll + (s[j] - '0')) % prime[k];
      }
    a[i] = now;
    for (int k = 0; k < l; ++k) mul[k] = (ll)mul[k] * a[i][k] % prime[k];
  }
  for (int i = 1; i <= n; ++i) {
    bool fail = 0;
    for (int k = 0; k < l; ++k)
      if (Qpow(a[i][k], n, prime[k]) != mul[k]) {
        fail = 1;
        break;
      }
    if (!fail) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}