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;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int v[SIZE];
int tt;
map<int, int> sg;
int solve(int x) {
  if (sg.count(x)) return sg[x];
  int res = 0, mask = 0, tmp[32] = {};
  for (int i = 0; (1 << i) <= x; i++) {
    tmp[solve((x & mask) | (x >> (i + 1)))] = 1;
    mask |= 1 << i;
  }
  while (tmp[res]) res++;
  return sg[x] = res;
}
int main() {
  sg[0] = 0;
  map<int, int> H;
  int(n);
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    int(x);
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int cnt = 0;
        while (x % j == 0) {
          cnt++;
          x /= j;
        }
        H[j] |= 1 << (cnt - 1);
      }
    }
    if (x != 1) H[x] |= 1;
  }
  int res = 0;
  for (auto it = H.begin(); it != H.end(); it++) {
    res ^= solve(it->second);
  }
  puts(res ? "Mojtaba" : "Arpa");
  return 0;
}
