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 n, ans;
map<long long, long long> sg, s;
void work(long long x) {
  long long tmp = x;
  for (long long i = 2; i * i <= tmp; i++)
    if (x % i == 0) {
      long long sum = 0;
      while (x % i == 0) sum++, x /= i;
      s[i] |= 1 << sum - 1;
    }
  if (x > 1) s[x] |= 1;
}
long long grundy(long long x) {
  if (x == 0) return 0;
  if (sg.count(x)) return sg[x];
  map<long long, long long> vis;
  long long tmp = x, m = 0;
  while (tmp) m++, tmp >>= 1;
  for (long long i = 1; i <= m; i++) {
    long long y = x >> i | (x & (1 << i - 1) - 1);
    vis[grundy(y)] = 1;
  }
  for (long long i = 0;; i++)
    if (!vis.count(i)) return sg[x] = i;
}
signed main() {
  cin >> n;
  for (long long i = 1, x; i <= n && cin >> x; i++) work(x);
  for (map<long long, long long>::iterator it = s.begin(); it != s.end(); it++)
    ans ^= grundy((*it).second);
  puts(ans ? "Mojtaba" : "Arpa");
  return 0;
}
