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;
const int maxn = 105;
map<int, int> mp, sg;
int dfs(int sta) {
  if (sta == 1) return 0;
  if (sg[sta]) return sg[sta];
  int bound, i;
  for (i = 1; i < 30; i++) {
    if (sta & (1 << i)) bound = i;
  }
  int vis = 0;
  for (i = 1; i <= bound; i++) {
    int tmp = (sta >> i) | (sta & ((1 << i) - 1));
    vis |= 1 << dfs(tmp);
  }
  int res = 0;
  for (i = 30; i >= 0; i--) {
    if ((~vis) & (1 << i)) res = i;
  }
  return sg[sta] = res;
}
int main() {
  int n, i, j, x;
  cin >> n;
  for (i = 0; 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++;
        }
        int v;
        if (!mp[j])
          v = 0;
        else
          v = mp[j];
        v |= 1 << cnt;
        mp[j] = v;
      }
    }
    if (x > 1) {
      int v;
      if (!mp[x])
        v = 0;
      else
        v = mp[x];
      v |= 1 << 1;
      mp[x] = v;
    }
  }
  int ans = 0;
  map<int, int>::iterator it;
  for (it = mp.begin(); it != mp.end(); it++) {
    ans ^= dfs(it->second);
  }
  if (ans)
    printf("Mojtaba");
  else
    printf("Arpa");
  return 0;
}
