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[100];
vector<int> primes;
map<int, int> mp, memo;
int mex(set<int> &vs) {
  int i = 0;
  while (true) {
    if (vs.find(i) == vs.end()) return i;
    i++;
  }
}
int f(int b) {
  if (memo.find(b) != memo.end()) return memo[b];
  set<int> vs;
  for (int k = 1; k <= 28; k++) {
    if ((b >> (k - 1)) == 0) break;
    vs.insert(f((b >> k) | (b & ((1 << (k - 1)) - 1))));
  }
  return memo[b] = mex(vs);
}
bool is_prime(int x) {
  if (x == 1) return false;
  for (int i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return false;
  }
  return true;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i <= 31622; i++) {
    if (is_prime(i)) primes.push_back(i);
  }
  cin >> N;
  for (int i = 0; i < (N); i++) cin >> A[i];
  for (int i = 0; i < (N); i++) {
    int x = A[i];
    for (int p : primes) {
      int ctr = 0;
      while (x % p == 0) x /= p, ctr++;
      if (ctr > 0) mp[p] |= 1 << (ctr - 1);
    }
    if (x > 1) mp[x] |= 1;
  }
  int g = 0;
  for (auto pp : mp) {
    int b = pp.second;
    g ^= f(b);
  }
  cout << (g == 0 ? "Arpa" : "Mojtaba") << "\n";
  return 0;
}
