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;
vector<pair<int, int> > factorize(int x) {
  vector<pair<int, int> > v;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      int cnt = 0;
      while (x % i == 0) x /= i, cnt++;
      v.emplace_back(i, cnt);
    }
  }
  if (x > 1) v.emplace_back(x, 1);
  return v;
}
map<int, int> dp;
int solve(int msk) {
  if (msk == 0) return 0;
  auto it = dp.find(msk);
  if (it != dp.end()) return it->second;
  int mx = 0;
  for (int i = 1; i < 30; i++) {
    if (msk & (1 << i)) {
      mx = i;
    }
  }
  vector<int> fr(32, 0);
  for (int j = 1; j <= mx; j++) {
    int newmask = 0;
    for (int k = 1; k <= mx; k++) {
      if (msk & (1 << k)) {
        if (k < j)
          newmask |= 1 << k;
        else if (k > j)
          newmask |= 1 << (k - j);
      }
    }
    fr[solve(newmask)] = 1;
  }
  for (int i = 0; i < 32; i++) {
    if (fr[i] == 0) return dp[msk] = i;
  }
}
int main() {
  int n;
  cin >> n;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    vector<pair<int, int> > r = factorize(x);
    for (pair<int, int> p : r) {
      mp[p.first] |= 1 << p.second;
    }
  }
  int x = 0;
  for (pair<int, int> p : mp) x ^= solve(p.second);
  if (x)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
  return 0;
}
