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 maxn = 500010, inf = 1e9;
int n, x;
map<int, int> mp, pri;
template <typename T>
inline void read(T &k) {
  int f = 1;
  k = 0;
  char c = getchar();
  while (c < '0' || c > '9') c == '-' && (f = -1), c = getchar();
  while (c <= '9' && c >= '0') k = k * 10 + c - '0', c = getchar();
  k *= f;
}
int dfs(int x) {
  if (!x) return 0;
  if (mp.count(x)) return mp[x];
  int p = x, cnt = 0;
  while (p) p >>= 1, cnt++;
  bool v[cnt + 1];
  for (int i = 0; i <= cnt; i++) v[i] = 0;
  for (int i = 1; i <= cnt; i++) {
    int tmp = dfs((x & ((1 << (i - 1)) - 1)) | (x >> i));
    if (tmp <= cnt) v[tmp] = 1;
  }
  for (int i = 0;; i++)
    if (!v[i]) return mp[x] = i;
}
inline void getpri(int x) {
  for (int i = 2; 1ll * i * i <= x; i++)
    if (x % i == 0) {
      int cnt = 0;
      while (x % i == 0) x /= i, cnt++;
      pri[i] |= (1 << (cnt - 1));
    }
  if (x != 1) pri[x] |= 1;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(x), getpri(x);
  int ans = 0;
  for (auto it = pri.begin(); it != pri.end(); it++) ans ^= dfs(it->second);
  puts(ans ? "Mojtaba" : "Arpa");
}
