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> m, mem;
void f(int x) {
  for (int i = 2; i * i <= x; i++) {
    int ct = 0;
    while (x % i == 0) x /= i, ct++;
    if (ct) m[i] |= (1 << (ct - 1));
  }
  if (x > 1) m[x] |= (1 << 0);
}
int solve(int x) {
  if (mem.count(x)) return mem[x];
  int ct = 0, msk = 0;
  bool tmp[32];
  memset(tmp, false, sizeof(tmp));
  for (int i = 0; (1 << i) <= x; i++)
    tmp[solve((x & msk) | (x >> (i + 1)))] = true, msk |= (1 << i);
  while (tmp[ct]) ct++;
  return mem[x] = ct;
}
int main() {
  int x;
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> x, f(x);
  long long res = 0;
  for (pair<int, int> v : m) res ^= solve(v.second);
  if (res == 0)
    cout << "Arpa";
  else
    cout << "Mojtaba";
  return 0;
}
