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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 1e5 + 5;
int n;
map<int, int> mp, memo;
int solve(int mask) {
  if (memo.count(mask)) return memo[mask];
  set<int> st;
  for (int k = 1; k <= 30; k++) {
    int nmask = (mask >> k) | (((1 << k) - 1) & mask);
    if (nmask != mask) st.insert(solve(nmask));
  }
  int mex = 0;
  while (st.count(mex)) mex++;
  return memo[mask] = mex;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0, x; i < n; i++) {
    cin >> x;
    for (int j = 2; j * j <= x; j++) {
      int p = 0;
      while (x % j == 0) x /= j, p++;
      if (p) mp[j] |= (1 << p);
    }
    if (x > 1) mp[x] |= 2;
  }
  int xr = 0;
  for (auto &i : mp) xr ^= solve(i.second);
  xr ? cout << "Mojtaba" : cout << "Arpa";
  return 0;
}
