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> mp, sg;
int grundy(int x) {
  if (sg.find(x) != sg.end()) return sg[x];
  unordered_set<int> vis;
  for (int i = 1; i <= 30; i++) {
    int y = (x >> i) | (x & ((1 << i - 1) - 1));
    if (y != x) vis.insert(grundy(y));
  }
  int res = 0;
  while (vis.find(res) != vis.end()) res++;
  return sg[x] = res;
}
int main() {
  int n, x;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int t = 0;
        while (x % j == 0) {
          x /= j;
          ++t;
        }
        mp[j] |= 1 << t - 1;
      }
    }
    if (x > 1) mp[x] |= 1;
  }
  int ans = 0;
  sg[0] = 0;
  for (auto &p : mp) {
    ans ^= grundy(p.second);
  }
  puts(ans ? "Mojtaba" : "Arpa");
  return 0;
}
