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, x;
map<int, int> M;
map<int, int>::iterator it;
map<int, int> Q;
int dfs(int v) {
  if (Q.count(v)) return Q[v];
  int mx = 0;
  for (int i = 1; i <= 30; i++)
    if (v >> i & 1) mx = i;
  if (mx == 0) return 0;
  int tmp;
  set<int> s;
  for (int i = 1; i <= mx; i++) {
    tmp = v & ((1 << i) - 1);
    for (int j = i; j <= 30; j++)
      if (v >> j & 1) tmp |= 1 << j - i;
    s.insert(dfs(tmp));
  }
  for (int i = 0; 1; i++)
    if (!s.count(i)) {
      Q[v] = i;
      return i;
    }
}
int main() {
  scanf("%d", &n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (int j = 2; 1ll * j * j <= x; j++)
      if (x % j == 0) {
        int t = 0;
        while (x % j == 0) {
          x /= j;
          t++;
        }
        M[j] |= 1 << t;
      }
    if (x != 1) M[x] |= 2;
  }
  for (it = M.begin(); it != M.end(); it++) {
    ans ^= dfs(it->second);
  }
  if (ans)
    puts("Mojtaba");
  else
    puts("Arpa");
  return 0;
}
