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;
unordered_map<int, int> p;
map<int, int> dp;
int grundy(int x) {
  if (dp[x]) return dp[x];
  set<int> s;
  for (int i = 1; i <= 30; i++) {
    int a = ((x >> i));
    if (a) s.insert(grundy(a | (x & ((1 << i) - 1))));
  }
  int v = 0;
  while (s.count(v)) {
    v++;
  }
  dp[x] = v;
  return v;
}
int main() {
  std::ios::sync_with_stdio(false);
  ;
  int n;
  std::cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    std::cin >> x;
    for (int j = 2; j * j <= x; j++) {
      int ctr = 0;
      while (x % j == 0) {
        ctr++;
        x /= j;
      }
      if (ctr) p[j] |= (1 << ctr);
    }
    if (x != 1) {
      p[x] |= (2);
    }
  }
  int ans = 0;
  for (auto it : p) {
    ans ^= (grundy(it.second));
  }
  if (ans)
    std::cout << "Mojtaba" << endl;
  else {
    std::cout << "Arpa" << endl;
  }
  return 0;
}
