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;
map<vector<int>, int> Map;
int grundy(vector<int> vec) {
  if (Map[vec] >= 1) return Map[vec] - 1;
  bool flag = false;
  vector<int> I(33, 0);
  for (int i = 31; i >= 1; i--) {
    if (vec[i] >= 1) flag = true;
    if (flag == false) continue;
    vector<int> vec2(32, 0);
    for (int j = 0; j < 32; j++) {
      if (vec[j] == 0) continue;
      if (j < i)
        vec2[j] = 1;
      else
        vec2[j - i] = 1;
    }
    int Z = grundy(vec2);
    I[Z] = 1;
  }
  for (int i = 0; i <= 32; i++) {
    if (I[i] == 0) {
      Map[vec] = i + 1;
      return i;
    }
  }
  return 31;
}
long long N, A[109];
vector<long long> E;
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    int cx = A[i];
    for (int j = 2; j * j <= cx; j++) {
      if (cx % j == 0) {
        E.push_back(j);
        while (cx % j == 0) cx /= j;
      }
    }
    if (cx >= 2) E.push_back(cx);
  }
  sort(E.begin(), E.end());
  E.erase(unique(E.begin(), E.end()), E.end());
  int S = 0;
  for (int i = 0; i < E.size(); i++) {
    vector<int> vec(32, 0);
    for (int j = 0; j < N; j++) {
      int cnt = 0, cx = A[j];
      while (cx % E[i] == 0) {
        cx /= E[i];
        cnt++;
      }
      vec[cnt] = 1;
    }
    int K = grundy(vec);
    S ^= K;
  }
  if (S >= 1)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
  return 0;
}
