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 MAX = 1e5 + 5;
int n, k;
map<int, set<int> > mp;
map<int, int> ans;
int que(int x) {
  if (ans.find(x) != ans.end()) return ans[x];
  vector<int> v;
  for (int i = 1; i <= 30; ++i) {
    int y = (((1 << (i - 1)) - 1) & x) | (x >> i);
    if (y == x) break;
    v.push_back(que(y));
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  int now = 0;
  for (int i : v) {
    if (now != i) {
      return ans[x] = now;
    }
    ++now;
  }
  return ans[x] = now;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    for (int i = 2; i <= sqrt(x); ++i) {
      int cnt = 0;
      while (x % i == 0) {
        ++cnt;
        x /= i;
      }
      if (cnt) {
        mp[i].insert(cnt);
      }
    }
    if (x != 1) mp[x].insert(1);
  }
  int sg = 0;
  for (auto itm : mp) {
    int x = 0;
    for (auto it : itm.second) {
      x |= (1 << (it - 1));
    }
    sg ^= que(x);
  }
  if (sg)
    cout << "Mojtaba\n";
  else
    cout << "Arpa\n";
}
