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;
map<long long, long long> dp1;
long long grundy(long long x) {
  if (x == 0) return 0;
  if (dp1.count(x) != 0) return dp1[x];
  set<long long> mex;
  long long lim = 0;
  for (long long i = 0; i < 30; i++)
    if ((1 << i) & x) lim = i;
  for (long long i = 0; i <= lim; i++) {
    long long y = 0;
    for (long long j = 0; j < i; j++)
      if ((1 << j) & x) y |= (1 << j);
    for (long long j = i + 1; j < 30; j++)
      if ((1 << j) & x) y |= (1 << (j - i - 1));
    mex.insert(grundy(y));
  }
  for (long long i = 0; true; i++)
    if (mex.count(i) == 0) {
      dp1[x] = i;
      return i;
    }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  map<long long, long long> ps;
  for (long long i = 0; i < n; i++) {
    map<long long, long long> m;
    long long x;
    cin >> x;
    for (long long j = 2; j * j <= x; j++) {
      while (x % j == 0) {
        m[j]++;
        x /= j;
      }
    }
    if (x != 1) m[x]++;
    for (pair<long long, long long> p : m) ps[p.first] |= (1 << (p.second - 1));
  }
  long long res = 0;
  for (pair<long long, long long> p : ps) {
    res ^= grundy(p.second);
  }
  if (res == 0)
    cout << "Arpa\n";
  else
    cout << "Mojtaba\n";
}
