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;
inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
inline void read(int &x) {
  static char c;
  x = 0;
  int sgn = 0;
  for (c = get(); c < '0' || c > '9'; c = get())
    if (c == '-') sgn = 1;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
  if (sgn) x = -x;
}
const long long MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
vector<int> primes;
void initp() {
  bitset<40000> isp;
  isp.set();
  isp[0] = isp[1] = false;
  for (int i = 2; i < 200; i++)
    if (isp[i])
      for (int j = i * i; j < 40000; j += i) isp[j] = false;
  for (int i = 2; i < 40000; i++)
    if (isp[i]) primes.push_back(i);
}
map<int, set<int>> primeset;
void getprimes(int n) {
  int lim = sqrt(n);
  for (int i = 0; primes[i] <= lim; i++) {
    if (n % primes[i] == 0) {
      int curr = 0;
      while (n % primes[i] == 0) {
        n /= primes[i];
        curr++;
      }
      primeset[primes[i]].insert(curr);
      lim = sqrt(n);
    }
  }
  if (n > 1) {
    primeset[n].insert(1);
  }
}
map<set<int>, int> dp;
int grundy(set<int> curr) {
  if (curr.empty()) return 0;
  if (dp.count(curr)) return dp[curr];
  set<int> mem;
  set<int> next;
  int lim = *curr.rbegin();
  for (int i = 1; i <= lim; i++) {
    next.clear();
    for (int count : curr) {
      if (count > i)
        next.insert(count - i);
      else if (count < i)
        next.insert(count);
    }
    mem.insert(grundy(next));
  }
  int ans = 0;
  while (mem.count(ans)) ans++;
  return dp[curr] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  initp();
  int n, a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    getprimes(a);
  }
  int ans = 0;
  for (pair<int, set<int>> p : primeset) {
    ans ^= grundy(p.second);
  }
  if (ans)
    cout << "Mojtaba" << '\n';
  else
    cout << "Arpa" << '\n';
  return 0;
}
