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 = 100055;
int not_prime[N], prime[N / 10];
int t;
map<int, int> SG, mask;
void init(int NN) {
  for (int i = 2; i <= NN; i++) {
    if (!not_prime[i]) prime[++t] = i;
    for (int j = 1; j <= t && prime[j] * i <= NN; j++) {
      not_prime[prime[j] * i] = 1;
      if (!(i % prime[j])) break;
    }
  }
}
int cal_SG(int a) {
  if (!a) return 0;
  if (SG.count(a)) return SG[a];
  int t = a, cnt = 0;
  while (t) t >>= 1, cnt++;
  map<int, int> tmp;
  for (int(i) = (1); (i) <= (cnt); (i)++)
    tmp[cal_SG(a >> i | (a & ((1 << i - 1) - 1)))] = 1;
  t = 0;
  while (tmp[t]) t++;
  return SG[a] = t;
}
void solve(int a) {
  int t, cnt;
  for (int i = 1; prime[i] * prime[i] <= a; i++) {
    t = prime[i], cnt = 0;
    while (!(a % t)) a /= t, cnt++;
    if (cnt) mask[prime[i]] = mask[prime[i]] | (1 << (cnt - 1));
  }
  if (a != 1) mask[a] |= 1;
}
int main() {
  int n, a, ans = 0;
  init(100005);
  cin >> n;
  for (int(i) = (1); (i) <= (n); (i)++) cin >> a, solve(a);
  map<int, int>::iterator i;
  for (i = mask.begin(); i != mask.end(); i++) ans ^= cal_SG(i->second);
  if (ans == 0)
    cout << "Arpa";
  else
    cout << "Mojtaba";
  return 0;
}
