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;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int mn = 2e5 + 10;
int thres = 1 << 17;
map<int, int> dp;
map<int, int> m;
int solve(int x) {
  bitset<32> num;
  if (x & 1) x--;
  if (x == 0) return 0;
  if (dp.count(x)) return dp[x];
  for (int j = 1; 1LL << j <= x; j++) {
    num[solve((x >> j) | (x & ((1 << j) - 1)))] = 1;
  }
  int ans = -1;
  for (int j = 0; 1LL << j <= x; j++)
    if (!num[j]) {
      ans = j;
      break;
    }
  num.reset();
  assert(ans >= 0);
  return dp[x] = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  solve(323425);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int num = 0;
        while (x % j == 0) {
          x /= j;
          num++;
        }
        m[j] |= 1 << num;
      }
    }
    if (x != 1) m[x] |= 2;
  }
  int ans = 0;
  for (pair<int, int> p : m) ans ^= solve(p.second);
  if (ans)
    printf("Mojtaba");
  else
    printf("Arpa");
}
