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> s;
set<int> st;
const int lim = 100005;
int a[111], i, j, k, n, m, v[110001], p[110001], c[110001], ans, t;
int calc(int x) {
  int i, k, res;
  set<int> b;
  if (!x) return 0;
  if (s.count(x)) return s[x];
  b.clear();
  for (k = 1; (x >> k); k++)
    ;
  for (i = 0; i < k; i++) {
    b.insert(calc((x & ((1 << i) - 1)) | (x >> (i + 1))));
  }
  for (res = 0; b.count(res); res++)
    ;
  return s[x] = res;
}
int main() {
  s.clear();
  memset(v, 0, sizeof(v));
  memset(c, 0, sizeof(c));
  for (i = 2, p[0] = 0; i <= lim; i++) {
    if (!v[i]) p[++p[0]] = i;
    for (j = 1; j <= p[0] && p[j] * i <= lim; j++) {
      v[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++) {
    for (k = a[i], j = 1; k > 1 && p[j] * p[j] <= k; j++) {
      for (t = 0; k % p[j] == 0; k /= p[j]) t++;
      if (t) c[p[j]] |= (1 << (t - 1));
    }
    if (k > 1) {
      if (k <= lim)
        c[k] |= 1;
      else if (!st.count(k))
        st.insert(k);
    }
  }
  if (st.size() & 1)
    ans = calc(1);
  else
    ans = 0;
  for (i = 1; i <= lim; i++)
    if (c[i]) ans ^= calc(c[i]);
  puts(ans ? "Mojtaba" : "Arpa");
  return 0;
}
