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 T>
void xchg(T &a, T &b) {
  T c = a;
  a = b;
  b = c;
}
void _main();
int main() {
  _main();
  return 0;
}
map<int, unsigned long long> P;
map<unsigned long long, unsigned long long> SG;
unsigned long long sg(unsigned long long x) {
  if (SG.count(x)) return SG[x];
  bool v[33];
  memset((v), 0, sizeof((v)));
  for (int i = 1; i <= 32; ++i) {
    int t = (x >> i) | (((1 << (i - 1)) - 1) & x);
    if (t == x) break;
    v[sg(t)] = 1;
  }
  for (int i = 0; i < 33; ++i) {
    if (!v[i]) {
      SG[x] = i;
      return i;
    }
  }
  SG[x] = 33;
  return 33;
}
void _main() {
  int N;
  cin >> N;
  while (N--) {
    int x;
    cin >> x;
    for (int p = 2; p * p <= x; ++p) {
      if (x % p) continue;
      x /= p;
      int b = 1;
      while (x % p == 0) {
        b <<= 1;
        x /= p;
      }
      if (P.count(p))
        P[p] |= b;
      else
        P[p] = b;
    }
    if (x > 1) {
      if (P.count(x))
        P[x] |= 1;
      else
        P[x] = 1;
    }
  }
  SG[0] = 0;
  for (int i = 1; i <= 32; ++i) {
    SG[1 << (i - 1)] = i;
  }
  int ans = 0;
  for (const auto &e : P) {
    ans ^= sg(e.second);
  }
  cout << (ans ? "Mojtaba" : "Arpa") << endl;
}
