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;
const int N = 102;
int n, a[N];
set<int> primes;
map<int, int> dp;
void getPrimes(int x) {
  if (x % 2 == 0) primes.insert(2);
  while (x % 2 == 0) x /= 2;
  for (int i = 3; 1ll * i * i <= x; i += 2)
    if (x % i == 0) {
      primes.insert(i);
      while (x % i == 0) x /= i;
    }
  if (x > 2) primes.insert(x);
}
int grundy(int msk) {
  if (msk == 1) return 0;
  if (dp.count(msk)) return dp[msk];
  set<int> canReach;
  for (int i = (1); i <= (int)(30); ++i)
    if ((1 << i) <= msk)
      canReach.insert(grundy((msk >> i) | (msk & ((1 << i) - 1))));
  for (int i = (0); i <= (int)(1000000000); ++i)
    if (canReach.find(i) == canReach.end()) return dp[msk] = i;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n); ++i) scanf("%d", &a[i]), getPrimes(a[i]);
  int ans = 0;
  for (int p : primes) {
    int msk = 0;
    for (int i = (1); i <= (int)(n); ++i) {
      int x = a[i], cnt = 0;
      while (x % p == 0) ++cnt, x /= p;
      msk |= 1 << cnt;
    }
    ans ^= grundy(msk);
  }
  cout << (ans ? "Mojtaba" : "Arpa");
}
