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;
int n, a[105];
set<int> vp;
map<int, int> mpx;
void part(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      vp.insert(i);
      while (x % i == 0) x /= i;
    }
  if (x > 1) vp.insert(x);
}
int deg(int x, int p) {
  int re = 0;
  while (x % p == 0) ++re, x /= p;
  return re;
}
int dfs(int s) {
  if (s == 1) return 0;
  if (mpx.count(s)) return mpx[s];
  vector<int> vi(33);
  for (int i = 1; i <= 30; ++i) {
    int t = s & ((1 << i) - 1) | s >> i;
    if (t != s) vi[dfs(t)] = 1;
  }
  for (int i = 0;; ++i)
    if (!vi[i]) {
      return mpx[s] = i;
    }
}
int solve(int p) {
  mpx.clear();
  int s = 1;
  for (int i = 1; i <= n; ++i) s |= 1 << deg(a[i], p);
  return dfs(s);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], part(a[i]);
  int ans = 0;
  for (auto p : vp) ans ^= solve(p);
  cout << (ans ? "Mojtaba" : "Arpa") << endl;
}
