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;
vector<int> w;
map<int, int> pr;
map<int, int> g;
int gr(int mask) {
  if (mask == 0) return 0;
  if (g.count(mask)) return g[mask];
  unsigned can = 0, x;
  for (int j = 29; j >= 1; --j) {
    x = 0;
    for (int i = 29; i >= 0; --i) {
      if (mask >> i & 1) {
        if (i < j) x |= 1 << i;
        if (i > j) x |= 1 << i - j;
      }
    }
    if (x == mask) continue;
    x = gr(x);
    if (x >= 0 and x <= 29) can |= 1 << x;
  }
  for (x = 0; can >> x & 1; ++x)
    ;
  return g[mask] = x;
}
int main() {
  ios::sync_with_stdio(0);
  int n, x;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x;
    int j = 2;
    while (x > 1) {
      if (x % j == 0) {
        int c = 0;
        while (x % j == 0) x /= j, ++c;
        pr[j] |= 1 << c;
      }
      ++j;
      if (1ll * j * j > x) j = x;
    }
  }
  int xs = 0;
  for (auto& i : pr) {
    xs ^= gr(i.second);
  }
  if (xs == 0)
    cout << "Arpa\n";
  else
    cout << "Mojtaba\n";
  return 0;
}
