#include <bits/stdc++.h>
using namespace std;
map<int, int> mp, pw;
int grundy(int mask) {
  if (mp.count(mask)) return mp[mask];
  unordered_set<int> s;
  for (int i = 1; i < 30; i++) {
    int msk = (mask >> i) | (mask & ((1 << (i - 1)) - 1));
    if (msk != mask) s.insert(grundy(msk));
  }
  int ans = 0;
  while (s.count(ans)) ans++;
  mp[mask] = ans;
  return ans;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    for (int j = 2; j * j <= x; j++) {
      int cnt = 0;
      while (x % j == 0) x /= j, cnt++;
      if (cnt) pw[j] |= (1 << (cnt - 1));
    }
    if (x > 1) pw[x] |= 1;
  }
  int ans = 0;
  mp[0] = 0;
  for (auto tp : pw) ans ^= grundy(tp.second);
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### Prompt

In CPP, your task is to solve 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, pw;
int grundy(int mask) {
  if (mp.count(mask)) return mp[mask];
  unordered_set<int> s;
  for (int i = 1; i < 30; i++) {
    int msk = (mask >> i) | (mask & ((1 << (i - 1)) - 1));
    if (msk != mask) s.insert(grundy(msk));
  }
  int ans = 0;
  while (s.count(ans)) ans++;
  mp[mask] = ans;
  return ans;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    for (int j = 2; j * j <= x; j++) {
      int cnt = 0;
      while (x % j == 0) x /= j, cnt++;
      if (cnt) pw[j] |= (1 << (cnt - 1));
    }
    if (x > 1) pw[x] |= 1;
  }
  int ans = 0;
  mp[0] = 0;
  for (auto tp : pw) ans ^= grundy(tp.second);
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```