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<long long, int> M;
int gn(long long I) {
  unordered_map<long long, int>::iterator it = M.find(I);
  if (it != M.end()) {
    return it->second;
  }
  bool B[64] = {0};
  for (int i = 0; i < 64; i++) {
    if ((I >> i) == 0) break;
    long long J = (I & ((1ll << i) - 1)) | (I >> (i + 1));
    B[gn(J)] = 1;
  }
  int ret = 0;
  while (B[ret]) ret++;
  M[I] = ret;
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &(a[i]));
  }
  vector<int> P;
  for (int i = 0; i < n; i++) {
    int N = a[i];
    for (int j = 2; j * j <= N; j++) {
      if (N % j == 0) {
        P.push_back(j);
        while (N % j == 0) {
          N /= j;
        }
      }
    }
    if (N > 1) {
      P.push_back(N);
    }
  }
  sort(P.begin(), P.end());
  int g = 0;
  for (int t = 0; t < P.size(); t++) {
    if (t > 0 && P[t - 1] == P[t]) continue;
    int p = P[t];
    long long I = 0;
    for (int i = 0; i < n; i++) {
      int k = 0, N = a[i];
      while (N % p == 0) {
        k++;
        N /= p;
        ;
      }
      if (k > 0) {
        I |= 1 << (k - 1);
      }
    }
    g ^= gn(I);
  }
  puts(g == 0 ? "Arpa" : "Mojtaba");
  return 0;
}
