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;
unordered_map<int, int> mp;
int dp(int msk) {
  if (mp.find(msk) != mp.end()) return mp[msk];
  set<int> st;
  for (int i = 0; (msk) >> i; i++) {
    int m = (msk & ((1 << (i)) - 1));
    m |= ((msk) >> (i + 1));
    st.insert(dp(m));
  }
  int ret = 0;
  while (!st.empty() && (*st.begin()) == ret) ret++, st.erase(*st.begin());
  return mp[msk] = ret;
}
int main() {
  int n, x, res = 0;
  scanf("%d", &n);
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++)
      if (!(x % j)) {
        int cnt = 0;
        while (!(x % j)) cnt++, x /= j;
        v.push_back({j, cnt - 1});
      }
    if (x != 1) v.push_back({x, 0});
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    int x = v[i].first, m = 0;
    while (i < v.size() && v[i].first == x) m |= (1 << (v[i].second)), i++;
    i--;
    res ^= dp(m);
  }
  if (res)
    puts("Mojtaba");
  else
    puts("Arpa");
  return 0;
}
