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 D = 40;
int n, sol;
map<int, int> ma, g, so;
int gam(int x) {
  if (x & 1) x--;
  if (!x) return 0;
  if (g[x]) return so[x];
  int p = 0, y = x;
  while (y > 1) y >>= 1, p++;
  int f[D];
  for (int i = 0; i < D; i++) f[i] = 0;
  for (int i = 1; i <= p; i++) f[gam((x & ((1 << i) - 1)) | (x >> i))] = 1;
  p = 0;
  while (f[p]) p++;
  g[x] = true;
  return so[x] = p;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++)
      if (!(x % j)) {
        int c = 0;
        while (!(x % j)) c++, x /= j;
        ma[j] |= 1 << c;
      }
    if (x > 1) ma[x] |= 2;
  }
  for (auto i : ma) sol ^= gam(i.second);
  if (sol)
    printf("Mojtaba");
  else
    printf("Arpa");
  return 0;
}
