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<int, int> r;
int gr(int x) {
  if (r.count(x)) return r[x];
  set<int> S;
  for (int i = 30; i; i--) {
    int lf = x >> i;
    if (lf > 0) S.insert(gr((x & ((1 << i) - 1)) | lf));
  }
  int m = 0;
  while (S.count(m)) m++;
  return r[x] = m;
}
int main() {
  int n, t, res = 0;
  cin >> n;
  map<int, int> pp;
  while (n--) {
    cin >> t;
    for (int i = 2; i * i <= t; i++) {
      int k = 0;
      while (t % i == 0) t /= i, k++;
      if (k) pp[i] |= (1 << k);
    }
    if (t != 1) pp[t] |= 2;
  }
  for (auto it : pp) res ^= gr(it.second);
  puts(res ? "Mojtaba" : "Arpa");
}
