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> ps;
int cal(int v) {
  static map<int, int> SG;
  SG[0] = -1;
  SG[1] = 1;
  auto &ans = SG[v];
  if (ans == 0) {
    set<int> nxt;
    for (int i = 0; i < 31; i++) {
      int d = v & ((1 << i) - 1) | (v >> (i + 1));
      if (d == v) break;
      nxt.insert(d);
    }
    map<int, bool> S;
    for (int x : nxt) S[cal(x)] = 1;
    for (int i = 0; i < 32; i++)
      if (S[i] == 0) {
        ans = i;
        break;
      }
    if (ans == 0) ans = -1;
  }
  if (ans == -1) return 0;
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0, x; i < n; i++) {
    scanf("%d", &x);
    for (int p = 2; p * p <= x; p++)
      if (x % p == 0) {
        int cnt = -1;
        while (x % p == 0) x /= p, cnt++;
        ps[p] |= (1 << cnt);
      }
    if (x != 1) ps[x] |= 1;
  }
  int ans = 0;
  for (auto &e : ps) ans ^= cal(e.second);
  puts(ans ? "Mojtaba" : "Arpa");
}
