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;
int N, n;
map<int, int> mask, G;
inline void add(int n) {
  int cnt;
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      cnt = 0;
      while (n % i == 0) n /= i, ++cnt;
      mask[i] |= 1 << cnt;
    }
  if (n > 1) mask[n] |= 1 << 1;
}
inline string bin(int m) {
  string res("00000000");
  for (int i = 0; i < m; ++i)
    if ((m >> i) & 1) res[i] = '1';
  reverse(res.begin(), res.end());
  return res;
}
inline int prin(int v, int m) { return v; }
int grundy(int Mask) {
  if (G.count(Mask)) return G[Mask];
  vector<int> v;
  for (int i = 1; i <= 30; ++i) {
    int x = (Mask >> i) | (Mask & ((1 << i) - 1));
    if (x == Mask) break;
    v.push_back(grundy(x));
  }
  sort(v.begin(), v.end());
  unique(v.begin(), v.end());
  for (int i = 0; i < v.size(); ++i)
    if (v[i] != i) return prin(G[Mask] = i, Mask);
  return prin(G[Mask] = v.size(), Mask);
}
void init() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    cin >> n;
    add(n);
  }
}
void sol() {
  int ans = 0;
  G[0] = G[1] = 0;
  for (auto x : mask) ans ^= grundy(x.second);
  if (ans)
    puts("Mojtaba");
  else
    puts("Arpa");
}
int main() {
  init();
  sol();
}
