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 MAXN = -1;
int n;
map<long long, int> Aux;
map<long long, set<int>> M;
void armame(long long x) {
  Aux.clear();
  for (long long d = 2; d * d <= x; d++) {
    while (x % d == 0) {
      Aux[d]++;
      x /= d;
    }
  }
  if (x > 1) Aux[x]++;
}
void metemos(long long x) {
  armame(x);
  for (auto it = Aux.begin(); it != Aux.end(); ++it) {
    long long p = it->first;
    int e = it->second;
    M[p].insert(e);
  }
}
map<long long, int> G;
int Grundy(long long bm) {
  if (bm == 1) return 0;
  if (G.count(bm)) return G[bm];
  int &ret = G[bm];
  vector<int> V(35, 0);
  for (int i = (1); i < (int)(35); i++) {
    long long nxt = (bm % (1 << i)) | (bm / (1 << i));
    if (nxt == bm) continue;
    V[Grundy(nxt)] = 1;
  }
  int c = 0;
  while (c < V.size() && V[c]) c++;
  return ret = c;
}
int solve(set<int> &S) {
  long long bm = 0;
  for (auto it = S.begin(); it != S.end(); ++it) {
    int i = (*it);
    bm += (1 << i);
  }
  return Grundy(bm);
}
int main() {
  while (scanf("%d", &n) == 1) {
    M.clear();
    for (int i = (0); i < (int)(n); i++) {
      long long x;
      scanf("%lld", &x);
      metemos(x);
    }
    int ans = 0;
    for (auto s = M.begin(); s != M.end(); ++s) {
      int lst = 0;
      ans ^= solve(s->second);
    }
    cout << (ans ? "Mojtaba" : "Arpa") << endl;
  }
  return 0;
}
