#include <bits/stdc++.h>
using namespace std;
map<int, int> mp, sg;
int dfs(int mask) {
  if (!mask) return 0;
  if (sg.find(mask) != sg.end()) return sg[mask];
  int mx = 0;
  for (int i = 1; i < 30; i++)
    if (mask >> i & 1) mx = i;
  int vis = 0;
  for (int i = 1; i <= mx; i++) {
    int tmask = ((mask >> (i))) | (mask & ((1 << i) - 1));
    vis |= (1 << dfs(tmask));
  }
  int res = 0;
  for (int i = 30; i >= 0; i--)
    if (~vis >> i & 1) res = i;
  return sg[mask] = res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    for (int i = 2; i * i <= a; i++)
      if (a % i == 0) {
        int cnt = 0;
        while (a % i == 0) a /= i, cnt++;
        mp[i] |= (1 << cnt);
      }
    if (a > 1) mp[a] |= (1 << 1);
  }
  int sum = 0;
  for (auto itr = mp.begin(); itr != mp.end(); itr++) sum ^= dfs(itr->second);
  printf("%s\n", (sum == 0 ? "Arpa" : "Mojtaba"));
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
map<int, int> mp, sg;
int dfs(int mask) {
  if (!mask) return 0;
  if (sg.find(mask) != sg.end()) return sg[mask];
  int mx = 0;
  for (int i = 1; i < 30; i++)
    if (mask >> i & 1) mx = i;
  int vis = 0;
  for (int i = 1; i <= mx; i++) {
    int tmask = ((mask >> (i))) | (mask & ((1 << i) - 1));
    vis |= (1 << dfs(tmask));
  }
  int res = 0;
  for (int i = 30; i >= 0; i--)
    if (~vis >> i & 1) res = i;
  return sg[mask] = res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    for (int i = 2; i * i <= a; i++)
      if (a % i == 0) {
        int cnt = 0;
        while (a % i == 0) a /= i, cnt++;
        mp[i] |= (1 << cnt);
      }
    if (a > 1) mp[a] |= (1 << 1);
  }
  int sum = 0;
  for (auto itr = mp.begin(); itr != mp.end(); itr++) sum ^= dfs(itr->second);
  printf("%s\n", (sum == 0 ? "Arpa" : "Mojtaba"));
  return 0;
}

```