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;
unordered_map<int, int> maskForPrime;
unordered_map<int, int> g;
int solve(int x) {
  if (g.count(x) > 0) return g[x];
  long long mex = 0;
  for (int k = 1;; k++) {
    int z = x >> k;
    if (z == 0) break;
    int y = (x ^ (z << k)) | z;
    y ^= (y & 1);
    mex |= 1LL << solve(y);
  }
  int k = 0;
  while ((mex >> k) & 1) k++;
  g[x] = k;
  return k;
}
int main() {
  int n;
  scanf("%d", &n);
  while (n--) {
    int x;
    scanf("%d", &x);
    for (int y = 2; y * y <= x; y++) {
      if (x % y != 0) continue;
      int k = 0;
      while (x % y == 0) {
        x /= y;
        k++;
      }
      if (maskForPrime.count(y) == 0) maskForPrime[y] = 0;
      maskForPrime[y] |= 1 << k;
    }
    if (x > 1) {
      if (maskForPrime.count(x) == 0) maskForPrime[x] = 0;
      maskForPrime[x] |= 1 << 1;
    }
  }
  g[0] = 0;
  int ans = 0;
  for (pair<int, int> t : maskForPrime) ans ^= solve(t.second);
  if (ans)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
  return 0;
}
