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<int, int> grundy;
int solve(int mask) {
  if (!mask) return 0;
  if (!grundy.count(mask)) {
    vector<int> v;
    int k = 1;
    for (int i = 32 - __builtin_clz(mask); i; i--) {
      int u = (mask >> i) | (((1 << (i - 1)) - 1) & mask);
      v.push_back(solve(u));
    }
    sort(v.begin(), v.end());
    int mex = 0;
    for (auto x : v) {
      if (x == mex)
        mex++;
      else if (x > mex)
        break;
    }
    grundy[mask] = mex;
  }
  return grundy[mask];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, x;
  cin >> n;
  map<int, int> prime;
  for (int i = 0; i < n; i++) {
    cin >> x;
    int y = x;
    for (int j = 2; j * j <= x; j++) {
      int cnt = 0;
      while (y % j == 0) {
        cnt++;
        y /= j;
      }
      if (cnt) prime[j] |= (1 << (cnt - 1));
    }
    if (y > 1) prime[y] |= 1;
  }
  int res = 0;
  for (auto p : prime) res ^= solve(p.second);
  if (res)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
  return 0;
}
