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 N = 110;
int n, arr[N];
int po = 0, num = 0, x = 0;
vector<pair<int, int> > v;
map<int, int> dp;
int get(int num) {
  if (num == 0) return 0;
  if (dp.find(num) != dp.end()) return dp[num];
  dp[num] = 0;
  long long vis = 0;
  int tmp;
  for (int i = 0; (1 << i) <= num; i++) {
    tmp = (((1 << i) - 1) & num);
    tmp |= (num >> (i + 1));
    vis |= ((long long)1 << get(tmp));
  }
  while (vis > 0) {
    if ((vis & 1) == 0) break;
    dp[num]++;
    vis >>= 1;
  }
  return dp[num];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    for (int j = 2; j * j <= arr[i]; j++) {
      po = 0;
      while (arr[i] % j == 0) {
        po++;
        arr[i] /= j;
      }
      if (po != 0) {
        v.push_back(make_pair(j, po));
      }
    }
    if (arr[i] != 1) v.push_back(make_pair(arr[i], 1));
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < (int)v.size(); i++) {
    num |= (1 << (v[i].second - 1));
    if (i == (int)v.size() - 1 || v[i].first != v[i + 1].first) {
      x ^= get(num);
      num = 0;
    }
  }
  if (x == 0)
    puts("Arpa");
  else
    puts("Mojtaba");
  return 0;
}
