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> dp;
map<int, int> f;
map<int, int>::iterator it;
int sg(int x) {
  if (dp.count(x)) return dp[x];
  int tot = 0;
  set<int> s;
  int a[40];
  for (int k = 1; (x >> k) >= 1; k++) {
    int y = (x >> k) | (x & ((1 << k) - 1));
    a[++tot] = sg(y);
  }
  sort(a + 1, a + tot + 1);
  int now = 1;
  int ans = 0;
  if (a[1] != 0) return dp[x] = 0;
  for (int i = 1; i < tot; i++) {
    if (a[i] + 1 >= a[i + 1]) continue;
    return dp[x] = a[i] + 1;
  }
  return dp[x] = a[tot] + 1;
}
int main() {
  int n, i, j, x;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int cnt = 0;
        while (x % j == 0) {
          x /= j;
          cnt++;
        }
        f[j] |= (1 << cnt);
      }
    }
    if (x != 1) f[x] |= 2;
  }
  dp[0] = dp[1] = 0;
  int ans = 0;
  for (it = f.begin(); it != f.end(); it++) {
    ans ^= sg(it->second);
  }
  if (ans == 0)
    printf("Arpa\n");
  else
    printf("Mojtaba\n");
}
