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 = 105;
int n, ans;
int a[maxn];
set<int> v;
map<int, int> d[maxn], f;
void divi(int num, int id) {
  int x = num;
  for (int i = 2; i * i <= num; i++)
    if (x % i == 0) {
      int cnt = 0;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      d[id][i] = cnt;
      v.insert(i);
    }
  if (x != 1) {
    d[id][x]++;
    v.insert(x);
  }
}
int sg(int s) {
  if (f.count(s)) return f[s];
  bool p[40];
  memset(p, 0, sizeof(p));
  for (int k = 1; (1 << k) <= s; k++) {
    int t = (s >> k) | (s & ((1 << k) - 1));
    p[sg(t)] = 1;
  }
  for (int i = 0;; i++)
    if (!p[i]) return f[s] = i;
}
int cal(int num) {
  int s = 0;
  for (int i = 1; i <= n; i++) s |= 1 << d[i][num];
  return sg(s);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    divi(a[i], i);
  }
  for (auto &i : v) ans ^= cal(i);
  if (ans == 0)
    puts("Arpa");
  else
    puts("Mojtaba");
  return 0;
}
