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 M = 1e9 + 7;
vector<pair<int, int> > V;
map<int, int> dp;
int grund(int x) {
  if (dp.count(x)) return dp[x];
  int vis[33] = {};
  for (int i = 1; (1 << i) <= x; i++) {
    int y = x >> i | x & ((1 << i) - 1);
    if (y != x) vis[grund(y)] = 1;
  }
  int i;
  for (i = 0; vis[i]; i++)
    ;
  return dp[x] = i;
}
int main() {
  int n;
  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 == 0) {
        pair<int, int> t;
        t.first = j;
        t.second = 0;
        while (x % j == 0) {
          x /= j;
          t.second++;
        }
        V.push_back(t);
      }
    }
    if (x > 1) {
      pair<int, int> t(x, 1);
      V.push_back(t);
    }
  }
  sort(V.begin(), V.end());
  int ans = 0;
  for (int i = 0; i < V.size(); i++) {
    int j = i, mask = 1 << (V[i].second);
    while (j + 1 < V.size() && V[i].first == V[j + 1].first) {
      j++;
      mask |= 1 << (V[j].second);
    }
    if (j - i + 1 < n) mask |= 1;
    ans ^= grund(mask);
    i = j;
  }
  puts((ans > 0) ? "Mojtaba" : "Arpa");
  return 0;
}
