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 long long oo = 1000000000000000000;
const int N = 1000006;
long long v[N];
map<int, int> dp;
int mex(const set<int> &s) {
  int ans = 0;
  while (s.count(ans)) ans++;
  return ans;
}
int g(int bit) {
  if (dp.count(bit)) return dp[bit];
  int lg = 31 - __builtin_clz(bit);
  set<int> s;
  for (int i = 0; i <= lg; i++) {
    int q = 0;
    for (int j = 0; j <= lg; j++)
      if (bit & (1 << j)) {
        if (j < i) q |= 1 << j;
        if (j > i) q |= 1 << (j - i - 1);
      }
    s.insert(g(q));
  }
  return dp[bit] = mex(s);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", v + i);
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    long long x = v[i];
    for (long long j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int tmp = 0;
        while (x % j == 0) tmp++, x /= j;
        m[j] |= 1 << (tmp - 1);
      }
    }
    if (x > 1) m[x] |= 1;
  }
  dp[0] = 0;
  int ans = 0;
  for (auto x : m) ans ^= g(x.second);
  printf("%s\n", ans ? "Mojtaba" : "Arpa");
  return 0;
}
