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;
long long ara[105], chk[100005];
map<long long, long long> mp, chkk, chkkk, value;
vector<long long> prime, sigh;
void setPrime() {
  long long i, j, k;
  for (i = 4; i < 100005; i += 2) chk[i] = 1;
  for (i = 3; i < 100005; i += 2)
    if (!chk[i])
      for (j = i * i; j < 100005; j += i) chk[j] = 1;
  for (i = 2; i < 100005; ++i)
    if (!chk[i]) prime.push_back(i);
}
long long grundy(long long mask) {
  if (!mask) return 0;
  if (chkk[mask]) return value[mask];
  set<long long> s;
  long long tmp;
  for (long long i = 1; i < 32; ++i) {
    tmp = (mask >> i) | (mask & ((1LL << (i - 1)) - 1LL));
    if (tmp < mask) s.insert(grundy(tmp));
  }
  tmp = 0;
  while (s.count(tmp)) tmp++;
  chkk[mask] = 1;
  return value[mask] = tmp;
}
int main() {
  setPrime();
  long long i, j, k;
  long long n, cnt, x, sz = prime.size(), sm;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> ara[i];
  for (i = 0; i < n; ++i) {
    x = ara[i];
    for (j = 0; j < sz; ++j) {
      cnt = 0;
      while (x % prime[j] == 0) x /= prime[j], cnt++;
      if (cnt) mp[prime[j]] |= (1LL << (cnt - 1));
    }
    if (x > 1) mp[x] |= (1LL);
    if (!chkkk[x] && x > 1) sigh.push_back(x), chkkk[x] = 1;
  }
  for (i = 0, sm = 0; i < sz; ++i) {
    sm ^= grundy(mp[prime[i]]);
  }
  for (i = 0; i < sigh.size(); ++i) sm ^= grundy(mp[sigh[i]]);
  if (sm)
    cout << "mojtAbA" << endl;
  else
    cout << "ArpA" << endl;
}
