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;
const int N = 1e5 + 5;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
int vis[N];
vector<int> prime;
void Sieve(int n) {
  for (int i = 2; i <= n; ++i)
    if (!vis[i]) {
      prime.push_back(i);
      for (int j = i + i; j <= n; j += i) vis[j] = 1;
    }
}
map<int, int> SG;
int dfs(int x) {
  if (!x)
    return 0;
  else if (SG.count(x))
    return SG[x];
  set<int> S;
  for (int i = 0; i < 32; ++i)
    if (x >> i) S.insert(dfs((x >> (i + 1)) | (x & ((1 << i) - 1))));
  for (int i = 0;; ++i)
    if (!S.count(i)) return SG[x] = i;
}
map<int, set<int> > M;
int main() {
  int n = read();
  Sieve(1e5);
  for (int i = 1; i <= n; ++i) {
    int x = read();
    for (int p : prime) {
      if (x % p == 0) {
        int cnt = 0;
        while (x % p == 0) ++cnt, x /= p;
        M[p].insert(cnt);
      }
      if (x == 0) break;
    }
    if (x > 1) M[x].insert(1);
  }
  int ans = 0;
  for (auto S : M) {
    int temp = 0;
    for (int i : S.second) temp |= (1 << (i - 1));
    ans ^= dfs(temp);
  }
  puts(ans ? "Mojtaba" : "Arpa");
  return 0;
}
