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 = 1e6 + 10;
const int SQRTN = 320;
const int LOGN = 20;
const double PI = acos(-1);
int n, val;
map<int, int> mp, dp;
map<int, int>::iterator it;
int grundy(int n) {
  if (dp.find(n) != dp.end()) return dp[n];
  int& mex = dp[n];
  set<int> st;
  int i = 1, nv = (n >> i) | (n & (1 << i) - 1);
  while (nv != n) {
    st.insert(grundy(nv));
    i++;
    nv = (n >> i) | (n & (1 << i) - 1);
  }
  while (st.find(mex) != st.end()) mex++;
  return mex;
}
void calc(int a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i) continue;
    int c = 0;
    while (a % i == 0) c++, a /= i;
    mp[i] |= 1LL << (c);
  }
  if (a > 1) mp[a] |= 1 << 1;
}
int main() {
  cin >> n;
  for (int a, i = 0; i < n; i++) scanf("%d", &a), calc(a);
  for (it = mp.begin(); it != mp.end(); ++it) val ^= grundy(it->second);
  if (val)
    cout << "Mojtaba" << endl;
  else
    cout << "Arpa" << endl;
}
