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 long double PI = acos(-1);
const int INF = 2e9 + 5;
const long long LINF = 1e18 + 5;
const long long MOD = 1e9 + 7;
const int MAX = 1e5 + 1;
bool prime[MAX];
long long primos[MAX];
int cp = 0;
int a[128];
int sol[32 * 100];
map<int, int> mapa;
int gid = 0;
int Get(int u) {
  if (mapa.count(u)) return mapa[u];
  mapa[u] = gid;
  gid++;
  return gid - 1;
}
map<int, int> dp;
int solve(int u) {
  if (u <= 1) return dp[u] = 0;
  if (dp.count(u)) return dp[u];
  int idx = -1;
  for (int i = 30; i >= 0; --i) {
    if ((1 << i) <= u) {
      idx = i;
      break;
    }
  }
  int v = u;
  int w = 0;
  int t = 0;
  int sh = idx;
  set<int> save;
  while (idx != 0) {
    v -= u & (1 << idx);
    t += u & (1 << idx);
    w = v | (t >> sh);
    save.insert(solve(w));
    --idx;
    --sh;
  }
  for (int i = 0;; ++i) {
    if (!save.count(i)) return dp[u] = i;
  }
}
int y[32];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  prime[0] = prime[1] = 1;
  for (int i = 2; i * i < MAX; ++i) {
    if (!prime[i])
      for (int j = i * i; j < MAX; j += i) {
        prime[j] = 1;
      }
  }
  for (int i = 0; i < MAX; ++i) {
    if (!prime[i]) {
      primos[cp++] = i;
    }
  }
  for (int i = 0; i < n; ++i) {
    int u = a[i];
    for (int j = 0; j < cp; ++j) {
      if (primos[j] * primos[j] > u) break;
      if (u % primos[j] == 0) {
        int ex = 0;
        while (u % primos[j] == 0) {
          ex++;
          u /= primos[j];
        }
        sol[Get(primos[j])] |= (1 << ex);
      }
    }
    if (u != 1) {
      sol[Get(u)] |= (1 << 1);
    }
  }
  long long ans = 0;
  for (int i = 0; i < 32 * 100; ++i) {
    if (sol[i] != 0) {
      ans ^= solve(sol[i]);
    }
  }
  if (ans == 0) {
    cout << "Arpa" << endl;
  } else
    cout << "Mojtaba" << endl;
  return 0;
}
