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 = 4e4 + 5;
map<int, int> G;
int b[33], sg[MAXN], T, vis[33];
vector<int> V;
int get(int s) {
  if (G.count(s)) return G[s];
  if (s == 0) return 0;
  vector<int> R;
  for (int j = 0; j < 29; j++)
    if (s >= b[j]) R.push_back(get((s % b[j]) | (s / b[j + 1])));
  ++T;
  for (int i = 0; i < R.size(); i++) vis[R[i]] = T;
  for (int i = 0; i < 30; i++)
    if (vis[i] != T) return G[s] = i;
}
int main() {
  int n;
  scanf("%d", &n);
  b[0] = 1;
  for (int i = 1; i < 33; i++) b[i] = b[i - 1] * 2;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 2; j < MAXN; j++) {
      int t = 0;
      while (x % j == 0) x /= j, t++;
      if (t) sg[j] |= b[t - 1];
    }
    if (x != 1) V.push_back(x);
  }
  sort(V.begin(), V.end());
  V.erase(unique(V.begin(), V.end()), V.end());
  int ans = 0;
  for (int j = 2; j < MAXN; j++) ans ^= get(sg[j]);
  ans ^= (int(V.size()) % 2);
  printf("%s\n", ans ? "Mojtaba" : "Arpa");
  return 0;
}
