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> bismarck, dp;
int n;
void factorization(int k) {
  for (int i = 2; i * i <= k; i++) {
    if (k % i == 0) {
      int c = 0;
      while (k % i == 0) {
        c++;
        k /= i;
      }
      dp[i] |= 1 << (c - 1);
    }
  }
  if (k > 1) {
    dp[k] |= 1;
  }
}
int grundy(int x) {
  if (x == 0) return 0;
  if (bismarck.count(x)) return bismarck[x];
  int s = 0;
  for (int i = 1, j = 1; i <= x; i <<= 1, j++) {
    s |= 1 << grundy((x >> j) | (x & ((1 << (j - 1)) - 1)));
  }
  int mex = 0;
  for (int i = 1; i <= s; i <<= 1, mex++) {
    if (!(s & i)) break;
  }
  return bismarck[x] = mex;
}
int main() {
  int k, ans = 0;
  cin >> n;
  while (n--) {
    cin >> k;
    factorization(k);
  }
  for (auto t : dp) {
    ans ^= grundy(t.second);
  }
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
}
