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> tmp, dp;
map<int, int>::iterator its;
int dfs(int state) {
  if (dp.count(state)) return dp[state];
  if (state == 0) return 0;
  set<int> cur;
  int res = 0;
  int tmp = 0, del = 0;
  for (int i = 30; i >= 0; i--) {
    if ((state >> i) & 1) del |= (1 << i);
    int nt = (state ^ del) | tmp;
    if (del == 0) continue;
    cur.insert(dfs(nt));
    tmp *= 2;
    if ((state >> i) & 1) tmp |= 1;
  }
  while (cur.find(res) != cur.end()) res++;
  return dp[state] = res;
}
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    int mx = a;
    for (int j = 2; j * j <= mx; j++) {
      int cnt = 0;
      while (a % j == 0) {
        cnt++;
        a /= j;
      }
      if (cnt > 0) tmp[j] |= (1 << (cnt - 1));
    }
    if (a != 1) tmp[a] |= 1;
  }
  int ans = 0;
  for (its = tmp.begin(); its != tmp.end(); its++) ans ^= dfs(its->second);
  if (ans > 0)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
}
