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;
int a;
map<int, int> prime, dp;
int dfs(int state) {
  if (state == 0) return 0;
  if (dp.count(state)) return dp[state];
  set<int> cur;
  int res = 0, tmp = 0, mx = 0;
  int mask = state;
  for (int i = 29; i >= 0; i--) {
    if ((state >> i) & 1) {
      mx = i;
      break;
    }
  }
  for (int i = 0; i <= mx; i++) {
    int nt = ((state >> (i + 1)) | tmp);
    cur.insert(dfs(nt));
    if ((state >> i) & 1) {
      tmp |= (1 << i);
      state ^= (1 << i);
    }
  }
  while (cur.find(res) != cur.end()) res++;
  return dp[mask] = res;
}
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    for (int j = 2; j * j <= a; j++) {
      if (a % j == 0) {
        int cnt = 0;
        while (a % j == 0) {
          cnt++;
          a /= j;
        }
        prime[j] |= (1 << (cnt - 1));
      }
    }
    if (a > 1) prime[a] |= 1;
  }
  int ans = 0;
  for (auto its : prime) {
    ans ^= dfs(its.second);
  }
  if (ans > 0)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
}
