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 = atan(1.0) * 4.0;
const long long int MOD = 1e9 + 7;
const long long int INF = 1e8;
const long double EPS = 0.0000001;
vector<int> primes;
void findP() {
  int m = 40000;
  bool arr[m];
  fill(arr, arr + m, 1);
  arr[0] = arr[1] = 0;
  for (int i = 2; i <= m; i++) {
    if (arr[i]) {
      primes.push_back(i);
      for (int j = i * i; j <= m; j += i) arr[j] = 0;
    }
  }
}
map<int, vector<int> > m1;
void solve(int x) {
  for (auto p : primes) {
    int cnt = 0;
    while (x % p == 0) cnt++, x /= p;
    if (cnt != 0) m1[p].push_back(cnt);
  }
  if (x != 1) m1[x].push_back(1);
}
map<int, int> m2;
int gnum(int n) {
  if (m2[n] != 0) return m2[n];
  int res = 0;
  vector<int> mex(30, 0);
  for (int i = 0; n >> i; i++) {
    int n1 = (n >> (i + 1)) | (n & ((1 << i) - 1));
    mex[gnum(n1)] = 1;
  }
  for (; mex[res]; res++)
    ;
  return (m2[n] = res);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  findP();
  int n, x;
  cin >> n;
  for (int i = 0; i <= n - 1; i++) {
    cin >> x;
    solve(x);
  }
  int ans = 0;
  for (auto E : m1) {
    int num = 0;
    for (auto v : E.second) num |= (1 << (v - 1));
    ans ^= gnum(num);
  }
  if (ans == 0)
    cout << "Arpa" << endl;
  else
    cout << "Mojtaba" << endl;
  return 0;
}
