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;
int a[101];
map<int, int> b;
map<int, int> mem;
int gr(int mask) {
  if (mem.count(mask) > 0) return mem[mask];
  int mex = 0;
  int k = 1;
  while (1) {
    int z = mask >> k;
    if (z == 0) break;
    int y = (mask ^ (z << k)) | z;
    y ^= (y & 1);
    mex |= 1 << gr(y);
    k++;
  }
  k = 0;
  while ((mex >> k) & 1) k++;
  return mem[mask] = k;
}
void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    for (int p = 2; p * p <= a[i]; p++) {
      int k = 0;
      while (a[i] % p == 0) {
        k++;
        a[i] /= p;
      }
      if (k) {
        b[p] |= 1 << k;
      }
    }
    if (a[i] > 1) {
      b[a[i]] |= 2;
    }
  }
}
int main() {
  init();
  int ans = 0;
  for (auto t : b) {
    ans ^= gr(t.second);
  }
  cout << (ans ? "Mojtaba" : "Arpa") << '\n';
}
