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 = 110;
int N, A[Maxn];
map<int, int> K, dp;
int SG(int x) {
  set<int> S;
  if (dp.count(x)) return dp[x];
  if (x == 0) return 0;
  for (int i = 1; i <= 30; i++) {
    int y = (x >> i) | (x & ((1 << (i - 1)) - 1));
    if (y != x)
      S.insert(SG(y));
    else
      break;
  }
  int p = 0;
  while (!S.empty()) {
    int now = *S.begin();
    if (now == p)
      p++;
    else if (now > p)
      return dp[x] = p;
    S.erase(S.begin());
  }
  return dp[x] = p;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= N; i++) {
    int x = A[i];
    for (int j = 2; j * j <= x; j++) {
      int k = 0;
      while (x % j == 0) {
        x /= j;
        k++;
      }
      if (k) K[j] |= (1 << (k - 1));
    }
    if (x != 1) K[x] |= 1;
  }
  int tmp = 0;
  dp[0] = 0;
  for (auto i : K) tmp ^= SG(i.second);
  if (tmp == 0)
    printf("Arpa\n");
  else
    printf("Mojtaba\n");
  return 0;
}
