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 N = 5000;
const int mx = 31;
int n, a[N];
map<int, int> dp;
map<int, vector<int> > q;
int step;
int used[5000000];
int solve(int mask) {
  if (dp.find(mask) != dp.end()) return dp[mask];
  if (mask == 0) {
    return dp[mask] = 0;
  }
  int maxVal = 0;
  for (int i = 1; i <= mx; i++) {
    if (mask & (1 << (i - 1))) maxVal = max(maxVal, i);
  }
  for (int i = 1; i <= maxVal; i++) {
    int mask1 = (mask & ((1 << (i - 1)) - 1));
    int mask2 = (mask >> i);
    solve(mask1 | mask2);
  }
  step++;
  int res = 0;
  for (int i = 1; i <= maxVal; i++) {
    int mask1 = (mask & ((1 << (i - 1)) - 1));
    int mask2 = (mask >> i);
    used[solve(mask1 | mask2)] = step;
  }
  while (used[res] == step) res++;
  return dp[mask] = res;
}
void proc(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      int p = i, c = 0;
      while (x % i == 0) {
        x /= i;
        c++;
      }
      q[p].push_back(c);
    }
  }
  if (x != 1) {
    q[x].push_back(1);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) proc(a[i]);
  vector<int> v;
  int ans = 0;
  for (auto it = q.begin(); it != q.end(); it++) {
    v.clear();
    for (int i = 0; i <= mx; i++) v.push_back(0);
    vector<int> &s = it->second;
    int mask = 0;
    for (int i = 0; i < s.size(); i++) {
      mask |= (1 << (s[i] - 1));
    }
    ans ^= solve(mask);
  }
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
  return 0;
}
