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;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
long long dp[1 << 20];
long long G(long long s) {
  if (s < (1 << 20)) return dp[s];
  long long ma = 0;
  for (long long i = 0; i < (33); i++)
    if (s >> i & 1) ma = i;
  bool ex[33] = {};
  for (long long i = 1; i <= ma + 1; i++) {
    long long k = s >> i;
    long long l = s & ((1ll << (i - 1)) - 1);
    long long g = G(k | l);
    ex[g] = true;
  }
  long long ret = 0;
  while (ex[ret]) ret++;
  return ret;
}
signed main() {
  dp[0] = 0;
  for (long long s = 1; s < (1 << 20); s++) {
    long long ma = 0;
    for (long long i = 0; i < (33); i++)
      if (s >> i & 1) ma = i;
    bool ex[33] = {};
    for (long long i = 1; i <= ma + 1; i++) {
      long long k = s >> i;
      long long l = s & ((1ll << (i - 1)) - 1);
      long long g = dp[k | l];
      ex[g] = true;
    }
    long long ret = 0;
    while (ex[ret]) ret++;
    dp[s] = ret;
  }
  long long N;
  cin >> N;
  vector<long long> A(N);
  for (long long i = 0; i < (N); i++) cin >> A[i];
  vector<long long> p;
  for (long long i = 0; i < (N); i++) {
    long long a = A[i];
    for (long long j = 2; j * j <= a; j++) {
      if (a % j) continue;
      p.push_back(j);
      while (a % j == 0) a /= j;
    }
    if (a != 1) p.push_back(a);
  }
  sort((p).begin(), (p).end());
  p.erase(unique((p).begin(), (p).end()), p.end());
  long long X = 0;
  for (long long i = 0; i < (p.size()); i++) {
    long long s = 0;
    for (long long j = 0; j < (A.size()); j++) {
      long long cnt = 0;
      while (A[j] % p[i] == 0) A[j] /= p[i], cnt++;
      if (cnt) s |= (1ll << (cnt - 1));
    }
    X ^= G(s);
  }
  if (X)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
  return 0;
}
