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 long double EPS = 1e-9, PI = acos(-1.);
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int N = 1e5 + 5;
long long k;
map<long long, long long> pr, gr;
void factorize(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    long long pot = 0;
    while ((n % i) == 0) {
      pot++;
      n /= i;
    }
    if (pot > 0) {
      if (!pr.count(i)) pr[i] = 0;
      pr[i] = pr[i] | (1ll << pot);
    }
  }
  if (n > 1) {
    if (!pr.count(n)) pr[n] = 0;
    pr[n] = pr[n] | (1ll << 1);
  }
}
long long grundy(long long mask) {
  if (mask <= 1) return 0;
  if (gr.count(mask)) return gr[mask];
  set<long long> possible;
  for (long long i = 1; (1ll << i) <= mask; i++) {
    long long now = mask & ((1ll << i) - 1);
    long long shift = (mask ^ now) >> i;
    possible.insert(grundy(now | shift));
  }
  long long ans = 0;
  for (auto it : possible) {
    if (it != ans) break;
    ans++;
  }
  return gr[mask] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> k;
  for (int i = 1; i <= k; i++) {
    long long x;
    cin >> x;
    factorize(x);
  }
  long long ans = 0;
  for (auto it : pr) {
    ans = ans ^ grundy(it.second);
  }
  cout << (ans ? "Mojtaba\n" : "Arpa\n");
  return 0;
}
