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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e2 + 10;
int a[N], n;
map<int, vector<int>> mp;
map<int, int> sg;
int SG(int x) {
  if (!x) return 0;
  if (sg.find(x) != sg.end()) return sg[x];
  int t[200];
  memset(t, 0, sizeof(t));
  bool f = false;
  for (int i = 30; i >= 0; --i) {
    if (x & (1 << i)) f = true;
    if (f) t[SG((x >> (i + 1)) | (x & ((1 << i) - 1)))] = 1;
  }
  sg[x] = 0;
  while (t[sg[x]]) sg[x]++;
  return sg[x];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j * j <= a[i]; ++j) {
      if (a[i] % j == 0) {
        int cnt = 0;
        while (a[i] % j == 0) a[i] /= j, ++cnt;
        mp[j].push_back(cnt);
      }
    }
    if (a[i] > 1) mp[a[i]].push_back(1);
  }
  int s = 0;
  for (auto it : mp) {
    int st = 0;
    for (int i = 0; i < it.second.size(); ++i) st |= 1 << it.second[i] - 1;
    s ^= SG(st);
  }
  if (s)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
  return 0;
}
