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;
int n, a[110], mx;
const int N = 30;
map<int, int> mem;
int Penu(int now) {
  if (!now || mem.find(now) != mem.end()) return mem[now];
  register int ans = 0;
  register set<int> mex;
  for (register int i = 1; i < 31; i++)
    if (now >> i) mex.insert(Penu(now & ((1 << i) - 1) | (now >> i)));
  while (mex.count(ans)) ans++;
  return mem[now] = ans;
}
int ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  mem[0] = 0;
  map<int, int> ma;
  for (register int i = 1, j; i <= n; i++) {
    cin >> j;
    for (register int r = 2, c = 0; r * r <= j; r++, c = 0) {
      while (!(j % r)) {
        c++;
        j /= r;
      }
      if (c) ma[r] |= (1 << c);
    }
    if (j > 1) ma[j] |= 2;
  }
  for (auto u : ma) ans ^= Penu(u.second);
  cout << (!ans ? "Arpa\n" : "Mojtaba\n");
  return 0;
}
