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 maxn = 200;
map<int, int> m, ma;
int n;
int a[maxn];
int calc(int val) {
  int &ans = m[val];
  if (ans == 0) {
    set<int> s;
    for (int i = 0; i <= 32; i++) {
      int num = (val & ((1 << i) - 1)) | (val >> (i + 1));
      if (num == val) break;
      s.insert(num);
    }
    int dd[40];
    memset(dd, 0, sizeof(dd));
    for (auto x : s) dd[calc(x)] = 1;
    for (int i = 0; i <= 32; i++)
      if (dd[i] == 0) {
        ans = i;
        break;
      }
    if (ans == 0) ans = -1;
  }
  if (ans == -1) return 0;
  return ans;
}
int main() {
  int ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 2; j <= sqrt(a[i]); j++) {
      if (a[i] % j) continue;
      int cnt = -1;
      while (a[i] % j == 0) {
        a[i] /= j;
        cnt++;
      }
      ma[j] |= (1 << cnt);
    }
    if (a[i] != 1) ma[a[i]] |= 1;
  }
  m[0] = -1;
  m[1] = 1;
  for (auto x : ma) ans ^= calc(x.second);
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
  return 0;
}
