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 MAXN = 105, mod = 1e9 + 7;
map<int, int> gru, mam;
int t[MAXN];
int rozw(int co) {
  if (gru.count(co)) return gru[co];
  int a, b, t[32];
  for (a = 0; a < 32; a++) t[a] = 0;
  for (a = 0; a < 31; a++)
    if (co >= (1 << a)) t[rozw((co >> (a + 1)) | (co % (1 << a)))] = 1;
  for (a = 0; t[a]; a++)
    ;
  gru[co] = a;
  return a;
}
int main() {
  int a, b, c, d, e, n, k, m, del, inc, w = 0;
  scanf("%d", &n);
  for (d = 0; d < n; d++) {
    scanf("%d", &e);
    b = e;
    for (a = 2; a * a <= e; a++) {
      for (c = 0; b % a == 0; c++) b /= a;
      if (c) mam[a] |= (1 << (c - 1));
    }
    if (b > 1) mam[b] |= 1;
  }
  for (auto x : mam) {
    a = x.second;
    w ^= rozw(a);
  }
  if (w)
    printf("mojtaba");
  else
    printf("arpa");
  return 0;
}
