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;
map<int, long long> G;
map<long long, int> SG;
int solve(long long n) {
  if (SG.find(n) != SG.end()) return SG[n];
  set<long long> Q;
  for (int x = 1; x < 32; x++) {
    long long t = n / (1LL << x);
    long long q = n % (1LL << x);
    if (t > 0) {
      long long r = (t | q);
      if (r % 2) r--;
      Q.insert(solve(r));
    }
  }
  for (int i = 0;; i++) {
    if (Q.find(i) == Q.end()) {
      SG[n] = i;
      return i;
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> R(N);
  for (int i = 0; i < N; i++) cin >> R[i];
  for (int i = 0; i < N; i++) {
    int k = R[i];
    for (int j = 2; j * j <= R[i]; j++) {
      int c = 0;
      while (k % j == 0) {
        c++;
        k /= j;
      }
      if (c != 0) G[j] |= (1LL << c);
    }
    if (k != 1) G[k] |= 2;
  }
  int v = 0;
  SG[0] = 0;
  for (auto it : G) {
    v ^= solve(it.second);
  }
  if (v == 0)
    cout << "Arpa" << endl;
  else
    cout << "Mojtaba" << endl;
  return 0;
}
