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<int, int> M, sg;
int dfs(int num) {
  if (!num) return 0;
  if (sg.find(num) != sg.end()) return sg[num];
  int mx;
  for (int i = 30; i >= 0; i--)
    if (num >> i & 1) {
      mx = i;
      break;
    }
  int vis = 0;
  int temp;
  for (int i = 1; i <= mx; i++) {
    temp = ((num >> (i + 1)) << 1) | (num & ((1 << i) - 1));
    vis |= (1 << dfs(temp));
  }
  for (int i = 0; i <= 30; i++)
    if ((~vis >> i) & 1) return sg[num] = i;
}
int main() {
  int n;
  cin >> n;
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++) {
      int cnt = 0;
      while (x % j == 0) {
        cnt++;
        x /= j;
      }
      if (cnt != 0) M[j] |= (1 << cnt);
    }
    if (x != 1) M[x] |= (1 << 1);
  }
  int sum = 0;
  for (auto it = M.begin(); it != M.end(); it++) {
    sum ^= dfs(it->second);
  }
  if (!sum)
    puts("Arpa");
  else
    puts("Mojtaba");
  return 0;
}
