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;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 0;
  for (; !isdigit(c); c = getchar()) f |= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ '0');
  if (f) x = -x;
}
template <typename T>
void write(T x, char ed = '\n') {
  static short st[30], tp;
  if (x < 0) putchar('-'), x = -x;
  do st[++tp] = x % 10, x /= 10;
  while (x);
  while (tp) putchar(st[tp--] | '0');
  putchar(ed);
}
int vis[50], n;
map<int, int> nm, mp;
int calc(int s) {
  if (nm.count(s)) return nm[s];
  for (int i = 1; 1 << i <= s; ++i) calc((s >> i) | (s & ((1 << i) - 1)));
  memset(vis, 0, sizeof(vis));
  for (int i = 1; 1 << i <= s; ++i)
    vis[calc((s >> i) | (s & ((1 << i) - 1)))] = 1;
  int x = 0;
  while (vis[x]) ++x;
  return nm[s] = x;
}
int main() {
  read(n), nm[1] = 0;
  for (int i = 1, x; i <= n; ++i) {
    read(x);
    for (int j = 2; j * j <= x; ++j)
      if (x % j == 0) {
        int s = 2;
        x /= j;
        while (x % j == 0) s <<= 1, x /= j;
        mp[j] |= s;
      }
    if (x != 1) mp[x] |= 2;
  }
  int nim = 0;
  for (auto t : mp) nim ^= calc(t.second);
  puts(nim ? "Mojtaba" : "Arpa");
  return 0;
}
