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> G;
long long grundy(long long msk) {
  if (G.find(msk) != G.end()) return G[msk];
  long long ret = 0;
  set<long long> vis;
  for (long long i = (1); i < (32); i++) {
    if (!(msk >> i)) break;
    long long nmsk = msk;
    nmsk &= ((1 << i) - 1);
    nmsk |= (msk >> i);
    long long j = 0;
    vis.insert(grundy(nmsk));
  }
  while (vis.find(ret) != vis.end()) ret++;
  return G[msk] = ret;
}
long long a[107];
map<long long, long long> p;
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = (0); i < (n); i++) {
    scanf("%lld", &a[i]);
    long long m = a[i];
    if (m % 2 == 0) {
      long long cnt = 0;
      while (m % 2 == 0) {
        cnt++;
        m /= 2;
      }
      p[2] |= (1 << cnt);
    }
    for (long long j = 3; (j * j) <= m; j += 2) {
      if ((m % j) == 0) {
        long long cnt = 0;
        while ((m % j) == 0) {
          cnt++;
          m /= j;
        }
        p[j] |= (1 << cnt);
      }
    }
    if (m > 1) {
      p[m] |= (1 << 1);
    }
  }
  long long val = 0;
  for (auto it = p.begin(); it != p.end(); it++) {
    val ^= grundy(it->second);
  }
  if (!val)
    printf("Arpa");
  else
    printf("Mojtaba");
}
