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 mod1 = 998244353, mod = 1e9 + 7;
const int MAX_N = 100005, MAX_M = 200010;
long long n, x;
map<long long, long long> dp, pp;
long long g(long long x) {
  if (dp.count(x)) return dp[x];
  unordered_set<long long> s;
  for (long long i = 1; i <= 30; ++i) {
    long long y = (x >> i) | (x & ((1 << (i - 1)) - 1));
    if (y != x) s.insert(g(y));
  }
  long long ans = 0;
  while (s.count(ans)) ++ans;
  return dp[x] = ans;
}
void Solve() {
  cin >> n;
  for (long long i = 0, x; i < n; i++) {
    cin >> x;
    for (long long p = 2; p * p <= x; ++p) {
      if (x % p == 0) {
        long long cn = 0;
        while (x % p == 0) {
          ++cn;
          x /= p;
        }
        pp[p] |= 1 << (cn - 1);
      }
    }
    if (x > 1) pp[x] |= 1;
  }
  dp[0] = 0;
  long long ans = 0;
  for (auto e : pp) ans ^= g(e.second);
  puts(ans ? "Mojtaba" : "Arpa");
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long TC = 1;
  while (TC--) Solve();
  return 0;
}
