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;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long exp(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b == 1) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
long long n;
map<long long, long long> dp;
long long g(long long mask) {
  if (dp.count(mask)) return dp[mask];
  if (mask <= 1) {
    return 0;
  }
  vector<long long> v;
  for (long long i = 32; i >= 1; i--) {
    long long go = (mask >> i) | (mask & ((1LL << i) - 1));
    if (go != mask) v.push_back(g(go));
  }
  sort(v.begin(), v.end());
  auto it = unique(v.begin(), v.end());
  v.resize(distance(v.begin(), it));
  long long mex = 0;
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] == mex) {
      mex++;
    } else
      break;
  }
  return dp[mask] = mex;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  vector<long long> v(n);
  map<long long, long long> m;
  for (long long &x : v) {
    cin >> x;
    if (x == 1) continue;
    for (long long i = 2; i * i <= x; i++) {
      if (x % i != 0) continue;
      long long j = 0;
      while (x % i == 0) {
        x /= i;
        j++;
      }
      m[i] |= (1LL << j);
    }
    if (x != 1) m[x] |= 2;
  }
  long long res = 0;
  for (auto p : m) {
    res ^= g(p.second);
  }
  if (res == 0) {
    cout << "Arpa"
         << "\n";
  } else
    cout << "Mojtaba"
         << "\n";
}
