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;
long long int n, a[105];
unordered_map<long long int, long long int> m;
unordered_map<long long int, long long int> dp;
long long int get(long long int msk) {
  if (msk == 0) return 0;
  if (dp.count(msk)) return dp[msk];
  set<long long int> s;
  for (int i = 1; i <= 32; i++)
    if ((msk >> i)) s.insert(get((msk >> i) | (msk & ((1 << i) - 1))));
  long long int x = 0;
  while (1) {
    auto j = s.find(x);
    if (j == s.end()) break;
    x++;
  }
  return dp[msk] = x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    long long int x = a[i];
    for (int j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        long long int cnt = 0;
        while (x % j == 0) {
          x /= j;
          cnt++;
        }
        m[j] = (m[j] | (1 << cnt));
      }
    }
    if (x >= 2) m[x] = (m[x] | 2);
  }
  long long int an = 0;
  for (auto j : m) {
    dp.clear();
    an = (an ^ get(j.second));
  }
  if (an == 0)
    cout << "Arpa" << '\n';
  else
    cout << "Mojtaba" << '\n';
  return 0;
}
