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;
using ll = long long;
int const nmax = 100;
int v[1 + nmax];
map<int, int> dp;
int mex(vector<int> v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < v.size(); i++)
    if (v[i] != i) return i;
  return v.size();
}
int computedp(int mask) {
  map<int, int>::iterator it = dp.find(mask);
  if (it != dp.end()) return it->second;
  vector<int> v;
  for (int i = 1; (1 << i) <= mask; i++) {
    int newmask = ((mask & ((1 << i) - 1)) | (mask >> i));
    v.push_back(computedp(newmask));
  }
  dp[mask] = mex(v);
  return dp[mask];
}
map<int, int> frec;
void decompose(int number) {
  for (int i = 2; i * i <= number; i++) {
    if (number % i == 0) {
      int p = 0;
      while (number % i == 0) {
        number /= i;
        p++;
      }
      frec[i] |= (1 << p);
    }
  }
  if (1 < number) frec[number] |= 2;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= n; i++) decompose(v[i]);
  int result = 0;
  for (map<int, int>::iterator it = frec.begin(); it != frec.end(); it++) {
    result ^= computedp(it->second | 1);
  }
  if (result == 0)
    cout << "Arpa";
  else
    cout << "Mojtaba";
  return 0;
}
