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<long long, long long> dp;
long long dfs(long long b) {
  if (b == 0) return 0;
  if (dp.count(b)) return dp[b];
  set<long long> s;
  for (long long i = 1; i < 32; i++) {
    long long mk = (1 << (i - 1)) - 1;
    long long nb = b & mk;
    nb |= (b >> i);
    if (nb == b) continue;
    s.insert(dfs(nb));
  }
  long long res = 0;
  while (s.count(res)) res++;
  return dp[b] = res;
}
signed main() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) {
    long long t = a[i];
    for (long long j = 2; j < 100000; j++) {
      if (t % j) continue;
      long long k = -1;
      while (t % j == 0) t /= j, k++;
      m[j] |= (1LL << k);
    }
    if (t != 1) m[t] |= 1;
  }
  long long grundy = 0;
  for (auto p : m) {
    grundy ^= dfs(p.second);
  }
  cout << (grundy ? "Mojtaba" : "Arpa") << endl;
  return 0;
}
