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<vector<int>, int> memo;
int grundy(vector<int> v) {
  if (v.empty()) return 0;
  if (memo.count(v)) return memo[v];
  vector<int> k;
  for (int i = 1; i <= v.back(); i++) {
    vector<int> ne;
    for (auto t : v) {
      if (t >= i) t -= i;
      if (t > 0) ne.push_back(t);
    }
    sort(ne.begin(), ne.end());
    ne.erase(unique(ne.begin(), ne.end()), ne.end());
    k.push_back(grundy(ne));
  }
  sort(k.begin(), k.end());
  k.erase(unique(k.begin(), k.end()), k.end());
  int ans = 0;
  while (ans < (int)k.size() && k[ans] == ans) ans++;
  return memo[v] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  map<int, set<int>> cnt;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    for (int j = 2; j * j <= a; j++) {
      if (a % j == 0) {
        int val = 0;
        while (a % j == 0) a /= j, val++;
        cnt[j].insert(val);
      }
    }
    if (a > 1) {
      cnt[a].insert(1);
    }
  }
  int res = 0;
  for (auto p : cnt) {
    vector<int> v;
    for (auto t : p.second) {
      v.push_back(t);
    }
    res ^= grundy(v);
  }
  cout << (res ? "Mojtaba" : "Arpa") << endl;
  return 0;
}
