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;
int n;
map<int, int> dp;
map<int, int> mask;
int remove(int x, int i) { return (x >> i) | (x & ((1 << (i - 1)) - 1)); }
int grundy(int x) {
  if (x == 0) return 0;
  if (dp.find(x) != dp.end()) return dp[x];
  bool mex[32];
  for (int i = 0; i < 32; i++) mex[i] = false;
  for (int i = 1; i < 32; i++) {
    if ((1 << (i - 1)) > x) break;
    mex[grundy(remove(x, i))] = true;
  }
  for (int i = 0; i < 32; i++)
    if (!mex[i]) return dp[x] = i;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    for (long long i = 2, cnt; i * i <= x; i++) {
      cnt = 0;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      if (cnt > 0) mask[i] |= (1 << (cnt - 1));
    }
    if (x > 1) mask[x] |= 1;
  }
  long long nim = 0;
  for (auto i : mask) {
    nim ^= grundy(i.second);
  }
  if (nim)
    cout << "Mojtaba\n";
  else
    cout << "Arpa\n";
}
