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;
inline void Fail() {
  printf("0");
  exit(0);
}
const int maxn = 105;
int n;
map<int, int> cnt, sg;
int grundy(int mask) {
  if (sg.count(mask)) return sg[mask];
  bool s[maxn] = {};
  for (int(i) = 0; (i) < 30; i++) {
    int nmask = mask & ((1 << i) - 1);
    nmask |= (mask >> i + 1);
    if (nmask == mask) break;
    s[grundy(nmask)] = true;
  }
  int res = 0;
  while (s[res]) res++;
  return sg[mask] = res;
}
int main() {
  scanf("%d", &n);
  for (int(i) = 0; (i) < n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++) {
      if (x % j) continue;
      int c = 0;
      while (x % j == 0) {
        c++;
        x /= j;
      }
      cnt[j] |= (1 << (c - 1));
    }
    if (x > 1) cnt[x] |= 1;
  }
  int ans = 0;
  for (map<int, int>::iterator it = cnt.begin(); it != cnt.end(); it++) {
    ans ^= grundy(it->second);
  }
  printf(ans ? "Mojtaba" : "Arpa");
  return 0;
}
