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> pr;
map<int, int> dp;
int dfs(int x) {
  if (dp.find(x) != dp.end()) return dp[x];
  int dpp[32] = {0};
  for (int i = 1; i <= x; i <<= 1) {
    dpp[dfs((x % i) | ((x / i) >> 1))] = 1;
  }
  for (int i = 0; i < 32; i++) {
    if (dpp[i] == 0) {
      dp[x] = i;
      return i;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int aa;
    scanf("%d", &aa);
    for (int j = 2; j * j <= aa; j++) {
      int cnt = 0;
      if (aa % j == 0) {
        while (aa % j == 0) {
          aa /= j;
          cnt++;
        }
        pr[j] |= (1 << (cnt - 1));
      }
    }
    if (aa > 1) {
      pr[aa] |= 1;
    }
  }
  int ans = 0;
  for (auto &target : pr) {
    dp.clear();
    dp[0] = 0;
    ans ^= dfs(target.second);
  }
  if (ans == 0)
    printf("Arpa\n");
  else
    printf("Mojtaba\n");
}
