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> prime, dp;
map<int, int>::iterator it;
int grundy(int mask) {
  if (mask <= 1) return dp[mask] = 0;
  if (dp[mask]) return dp[mask];
  vector<int> grnum(30, 0);
  int num = 0, j = 29, bits = 0;
  for (int i = 1; mask >> i; i++) {
    grnum[(grundy((mask >> i) | (mask & ((1 << i) - 1))))] = 1;
  }
  for (int i = 0; i < 30; i++) {
    if (grnum[i] == 0) {
      dp[mask] = i;
      break;
    }
  }
  return dp[mask];
}
int main() {
  int n, val, cnt, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val);
    for (int j = 2; j * j <= val; j++) {
      if (val % j) continue;
      cnt = 0;
      while (val % j == 0) {
        val /= j;
        cnt++;
      }
      prime[j] |= (1 << cnt);
    }
    if (val > 1) prime[val] |= 2;
  }
  for (it = prime.begin(); it != prime.end(); it++) {
    ans ^= grundy(it->second);
  }
  if (ans)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
  return 0;
}
