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 = 106;
int a[N];
map<long long, long long> mp;
int get_mex(vector<int> v) {
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; v.size() > i; i++) {
    if (v[i] != i) return i;
  }
  return v.size();
}
long long solve(long long x) {
  if (mp.find(x) != mp.end()) return mp[x];
  vector<int> v;
  for (int i = 1;; i++) {
    if (x < (1LL << (i - 1))) break;
    v.push_back(solve((x & ((1LL << (i - 1)) - 1) | (x >> i))));
  }
  return mp[x] = get_mex(v);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> v;
  for (int i = 1; n >= i; i++) {
    scanf("%d", &a[i]);
    int tmp = a[i];
    for (int j = 2; j * j <= tmp; j++) {
      if (tmp % j == 0) {
        v.push_back(j);
        while (tmp % j == 0) tmp /= j;
      }
    }
    if (tmp != 1) {
      v.push_back(tmp);
    }
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  mp[0] = 0;
  long long sg = 0;
  for (int p : v) {
    long long mask = 0;
    for (int i = 1; n >= i; i++) {
      int cnt = 0;
      while (a[i] % p == 0) {
        ++cnt;
        a[i] /= p;
      }
      if (cnt) {
        mask |= (1 << (cnt - 1));
      }
    }
    sg ^= solve(mask);
  }
  if (sg)
    puts("Mojtaba");
  else
    puts("Arpa");
}
