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 sg[1000006], a[1000], ans = 0;
bool vis[30], mark[1000006];
int SG(int x) {
  if (x < 1000006) return sg[x];
  bool vi[30] = {0};
  for (int k = 0; (1 << k) <= x; k++) {
    int y = ((x % (1 << k)) | (x >> (k + 1)));
    vi[SG(y)] = 1;
  }
  for (int i = 0; i < 30; i++) {
    if (vi[i] == 0) return i;
  }
}
int main() {
  for (int i = 1; i < 1000006; i++) {
    for (int k = 0; (1 << k) <= i; k++) {
      int x = ((i % (1 << k)) | (i >> (k + 1)));
      vis[sg[x]] = 1;
    }
    for (int k = 0; k < 30; k++) {
      if (vis[k] == 0) {
        sg[i] = k;
        break;
      }
    }
    memset(vis, 0, sizeof(vis));
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 2; i < 100005; i++) {
    if (mark[i]) continue;
    int x = 0;
    for (int k = i; k < 100005; k += i) mark[k] = 1;
    for (int k = 0; k < n; k++) {
      int cnt = 0;
      while (a[k] % i == 0) {
        a[k] /= i;
        cnt++;
      }
      if (cnt) x |= (1 << (cnt - 1));
    }
    ans ^= SG(x);
  }
  sort(a, a + n);
  int p = 1;
  for (int i = 0; i < n; i++) {
    if (a[i] != p) {
      ans ^= 1, p = a[i];
    }
  }
  if (ans == 0) {
    printf("Arpa\n");
  } else {
    printf("Mojtaba\n");
  }
}
