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 const maxn = 1e6 + 10;
int tot = 0;
int a[110];
int num[maxn];
unordered_map<int, int> vis, sg, sgv;
int dfs(int x) {
  if (sgv[x]) return sg[x];
  sgv[x] = 1;
  bool v[100] = {0};
  int pos = x & -x;
  for (int i = 0; (1 << i) <= x; i++) {
    v[dfs(x % (1 << i) | x / (1 << i + 1))] = 1;
  }
  for (int i = 0; i < 100; i++)
    if (!v[i]) return sg[x] = i;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 2; j * j <= a[i]; j++) {
      int k = 0;
      while (a[i] % j == 0) {
        k++;
        a[i] /= j;
      }
      if (!k) continue;
      int x = vis[j];
      if (x)
        num[x] |= (1 << (k - 1));
      else {
        num[++tot] |= (1 << (k - 1));
        vis[j] = tot;
      }
    }
    if (a[i] > 1) {
      int x = vis[a[i]];
      if (x)
        num[x] |= 1;
      else {
        num[++tot] |= 1;
        vis[a[i]] = tot;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= tot; i++) {
    ans ^= dfs(num[i]);
  }
  if (ans)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
}
