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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
map<int, int> memo;
int solve(int mask) {
  if (!mask) return 0;
  if (memo.find(mask) != memo.end()) return memo[mask];
  map<int, bool> canreach;
  for (int i = 0; i < 32; i++) {
    int aux = ((1 << i) - 1) & mask;
    int ns = aux | ((mask - aux) >> (i + 1));
    if (ns == mask) continue;
    canreach[solve(ns)] = true;
  }
  int mex = 0;
  for (int i = 0; i < 100; i++)
    if (!canreach[i]) {
      mex = i;
      break;
    }
  return memo[mask] = mex;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  map<int, int> mps;
  for (int j = 0; j < n; j++) {
    int val, aux;
    cin >> val;
    aux = val;
    for (int i = 2; i * i <= val; i++) {
      int cnt = 0;
      while (aux % i == 0) {
        aux /= i;
        cnt++;
      }
      if (cnt) mps[i] |= (1 << (cnt - 1));
    }
    if (aux != 1) mps[aux] |= 1;
  }
  int grundy = 0;
  for (auto u : mps) grundy ^= solve(u.second);
  cout << (grundy ? "Mojtaba" : "Arpa") << '\n';
  return 0;
}
