Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.

In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.

Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.

Output

If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).

You can print each letter in any case (upper or lower).

Examples

Input

4
1 1 1 1


Output

Arpa


Input

4
1 1 17 17


Output

Mojtaba


Input

4
1 1 17 289


Output

Arpa


Input

5
1 2 3 4 5


Output

Arpa

Note

In the first sample test, Mojtaba can't move.

In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].

In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
#include <bits/stdc++.h>
using namespace std;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "sz = " << vc.size() << endl << "[";
  for (const T& v : vc) o << v << ",";
  o << "]";
  return o;
}
const long long ccc = 32000;
bool prime[ccc + 1];
vector<long long> pr;
void makeprime() {
  long long i, j;
  for (i = 2; i <= ccc; i++) prime[i] = true;
  for (i = 2; i * i <= ccc; i++)
    if (prime[i])
      for (j = 2; j * i <= ccc; j++) prime[j * i] = false;
  for (i = 2; i <= ccc; i++)
    if (prime[i]) pr.push_back(i);
}
bool B(int x, int i) { return (x >> i) & 1; }
int N;
long long a[100];
map<int, int> mp;
int g(int x) {
  if (x <= 1) return 0;
  if (mp.count(x)) return mp[x];
  int nxt = 0;
  for (int i = 1; i <= (int)(30); i++) {
    if ((1 << i) > x) break;
    int y = (x % (1 << i)) | (x >> i);
    nxt |= 1 << g(y);
  }
  for (int i = 0; i < (int)(31); i++)
    if (!B(nxt, i)) {
      return mp[x] = i;
    }
  assert(0);
}
int main() {
  makeprime();
  cin >> N;
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  set<int> st;
  for (int i = 0; i < (int)(N); i++) {
    int x = a[i];
    for (int p : pr)
      if (x % p == 0) {
        while (x % p == 0) x /= p;
        st.insert(p);
      }
    if (x > 1) {
      st.insert(x);
    }
  }
  int res = 0;
  for (int p : st) {
    int b = 0;
    for (int i = 0; i < (int)(N); i++) {
      int x = a[i];
      int t = 0;
      for (; x % p == 0; t++) x /= p;
      b |= 1 << t;
    }
    res ^= g(b);
  }
  if (res)
    puts("Mojtaba");
  else
    puts("Arpa");
}
