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 grundy[(1 << 20)];
int masks[10000];
int mex(set<int> se) {
  int i = 0;
  set<int>::iterator it;
  for (it = se.begin(), i = 0; it != se.end(); ++it, i++) {
    if (*it != i) return i;
  }
  return i;
}
int cnt(int mask) {
  int i, j, k, p, q, x, y;
  set<int> se;
  if (mask == 0) return 0;
  if (mask < (1 << 20) && grundy[mask] != -1) return grundy[mask];
  for (i = 0; i < 31; i++) {
    p = mask >> (i + 1);
    q = mask & ((1 << i) - 1);
    x = p | q;
    if (x == mask) break;
    se.insert(cnt(x));
  }
  y = mex(se);
  if (mask < (1 << 20)) grundy[mask] = y;
  return y;
}
int main() {
  int n, i, j, k, p, q, x, y, cntt = 0;
  map<int, int> ma;
  memset(grundy, -1, sizeof(grundy));
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    for (j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        if (ma.find(j) == ma.end()) ma[j] = ++cntt;
        p = ma[j];
        q = -1;
        while (x % j == 0) {
          x /= j;
          q++;
        }
        masks[p] |= (1 << q);
      }
    }
    if (x != 1) {
      j = x;
      if (ma.find(j) == ma.end()) ma[j] = ++cntt;
      p = ma[j];
      masks[p] |= 1;
    }
  }
  x = 0;
  for (i = 1; i <= cntt; i++) x ^= cnt(masks[i]);
  printf("%s\n", x ? "Mojtaba" : "Arpa");
  return 0;
}
