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;
map<int, int> sg;
int cnt[40005][105], p[40005], tot, arr[105];
bool np[40005];
int work(int x) {
  if (x == 1) return 0;
  if (sg.count(x)) return sg[x];
  bool vis[105] = {};
  for (int i = 29; i; i--) {
    if (x >= (1 << i)) {
      int res = work((x >> i) | (x & ((1 << i) - 1)));
      vis[res] = true;
    }
  }
  for (int i = 0;; i++) {
    if (!vis[i]) return sg[x] = i;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", arr + i);
  for (int i = 2; i < 32005; i++) {
    if (!np[i]) p[tot++] = i;
    for (int j = 0; j < tot && i * p[j] < 32005; j++) {
      np[i * p[j]] = true;
      if (!(i % p[j])) break;
    }
  }
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    for (int j = 0; j < tot; j++) {
      while (!(x % p[j])) x /= p[j];
    }
    if (x != 1) p[tot++] = x;
  }
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    for (int j = 0; j < tot; j++) {
      while (!(x % p[j])) {
        x /= p[j];
        cnt[j][i]++;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < tot; i++) {
    int x = 0;
    for (int j = 0; j < n; j++) x |= 1 << cnt[i][j];
    sg.clear();
    ans ^= work(x);
  }
  printf("%s\n", !ans ? "Arpa" : "Mojtaba");
  return 0;
}
