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;
template <class P, class Q>
inline void smin(P &a, Q b) {
  if (b < a) a = b;
}
template <class P, class Q>
inline void smax(P &a, Q b) {
  if (a < b) a = b;
}
int n;
map<int, int> f;
map<int, int> m;
int nim(int r) {
  if ((r & (r - 1)) == 0) return r;
  if (m.find(r) != m.end()) return m[r];
  int mark = 0;
  for (int i = (int)(1), i__b = (int)(31); i < i__b; ++i) {
    if (r < (1 << i)) break;
    mark |= nim((r & ((1 << i) - 1)) | (r >> i));
  }
  mark = ~mark;
  return m[r] = mark & -mark;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0, i__n = (int)(n); i < i__n; ++i) {
    int a;
    cin >> a;
    for (int i = 2; i * i <= a; i++)
      if (a % i == 0) {
        int c = 0;
        while (a % i == 0) a /= i, c++;
        f[i] |= 1 << c;
      }
    if (a > 1) f[a] |= 1 << 1;
  }
  int x = 0;
  for (pair<int, int> p : f) x ^= __builtin_ctz(nim(p.second));
  cout << (x ? "Mojtaba" : "Arpa") << endl;
  return 0;
}
