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;
unordered_map<int, long long> pile;
unordered_map<int, unordered_map<long long, int>> dp;
int calc(int u, long long msk) {
  if (!msk || msk == 1) return 0;
  if (dp[u].find(msk) != dp[u].end()) return dp[u][msk];
  int &ref = dp[u][msk];
  bool arr[62] = {};
  int to = log(msk) / log(2) + 1e-9;
  for (int i = 1; i <= to; i++)
    arr[calc(u, (msk >> i) | (msk & ((1 << i)) - 1))] = 1;
  for (int i = 0; i < 62; i++)
    if (!arr[i]) return ref = i;
}
vector<int> v;
void find(int a) {
  for (long long i = 2; i * i <= a; i++) {
    int freq = 0;
    while (!(a % i)) {
      freq++;
      a /= i;
    }
    pile[i] |= 1 << freq;
    if (freq) v.push_back(i);
  }
  if (a != 1) pile[a] |= 1 << 1, v.push_back(a);
}
int main() {
  int n;
  scanf("%d", &n);
  int lel = 0;
  while (n--) {
    int a;
    scanf("%d", &a);
    find(a);
  }
  for (auto i : pile) lel ^= calc(i.first, i.second);
  puts(lel ? "Mojtaba" : "Arpa");
}
