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> p;
map<int, int> f;
int mex(int x) {
  if (f.count(x)) return f[x];
  set<int> S;
  for (int i = 1; (1 << i) <= x; ++i) {
    int y = (x >> i) | (x & ((1 << i) - 1));
    S.insert(mex(y));
  }
  for (int i = 0; i < x; ++i) {
    if (S.find(i) == S.end()) {
      f[x] = i;
      return i;
    }
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    for (int i = 2; i * i <= x; ++i) {
      if (x % i) continue;
      int cnt = 0;
      while (x % i == 0) {
        ++cnt;
        x /= i;
      }
      p[i] |= (1 << cnt);
    }
    if (x != 1) p[x] |= 2;
  }
  int ans = 0;
  f[1] = 0;
  for (auto x : p) {
    ans ^= mex(x.second);
  }
  if (!ans)
    puts("Arpa");
  else
    puts("Mojtaba");
  return 0;
}
