#include <bits/stdc++.h>
using namespace std;
map<int, int> dp;
map<int, int> prime;
int solve(int mask) {
  if (dp.count(mask)) return dp[mask];
  int &ret = dp[mask];
  set<int> st;
  for (int i = 1; i <= 30; i++) {
    int nw = 0;
    for (int j = 1; j <= 30; j++) {
      if (j < i && (mask & (1 << j)))
        nw |= (1 << j);
      else if (j > i && (mask & (1 << j)))
        nw |= (1 << (j - i));
    }
    if (nw != mask) st.insert(solve(nw));
  }
  int ans = 0;
  while (st.count(ans)) ans++;
  return ret = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  int cnt;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    for (long long j = 2; j * j <= a; j++) {
      if (a % j == 0) {
        cnt = 0;
        while (a % j == 0) {
          cnt++;
          a /= j;
        }
        prime[j] |= (1 << (cnt));
      }
    }
    if (a > 1) prime[a] |= 2;
  }
  for (map<int, int>::iterator it = prime.begin(); it != prime.end(); it++) {
    ans ^= solve(it->second);
  }
  if (ans != 0) {
    puts("Mojtaba");
  } else
    puts("Arpa");
  return 0;
}

### Prompt

Your challenge is to write 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> dp;
map<int, int> prime;
int solve(int mask) {
  if (dp.count(mask)) return dp[mask];
  int &ret = dp[mask];
  set<int> st;
  for (int i = 1; i <= 30; i++) {
    int nw = 0;
    for (int j = 1; j <= 30; j++) {
      if (j < i && (mask & (1 << j)))
        nw |= (1 << j);
      else if (j > i && (mask & (1 << j)))
        nw |= (1 << (j - i));
    }
    if (nw != mask) st.insert(solve(nw));
  }
  int ans = 0;
  while (st.count(ans)) ans++;
  return ret = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  int cnt;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    for (long long j = 2; j * j <= a; j++) {
      if (a % j == 0) {
        cnt = 0;
        while (a % j == 0) {
          cnt++;
          a /= j;
        }
        prime[j] |= (1 << (cnt));
      }
    }
    if (a > 1) prime[a] |= 2;
  }
  for (map<int, int>::iterator it = prime.begin(); it != prime.end(); it++) {
    ans ^= solve(it->second);
  }
  if (ans != 0) {
    puts("Mojtaba");
  } else
    puts("Arpa");
  return 0;
}

```