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;
constexpr int MAX = 128;
int a[MAX];
unordered_map<int, int> cnt, dp;
void Calc(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (!(x % i)) {
      int cur = 0;
      for (; !(x % i); x /= i) {
        ++cur;
      }
      cnt[i] |= (1 << (cur - 1));
    }
  }
  if (x > 1) {
    cnt[x] |= 1;
  }
}
int Mex(vector<int> &arr) {
  sort(arr.begin(), arr.end());
  arr.resize(unique(arr.begin(), arr.end()) - arr.begin());
  int pre = -1;
  for (int it : arr) {
    if (it != pre + 1) {
      return pre + 1;
    }
    pre = it;
  }
  return pre + 1;
}
int DFS(int x) {
  if (!x) return 0;
  auto it = dp.find(x);
  if (it != dp.end()) {
    return it->second;
  }
  int &ret = dp[x];
  vector<int> buf;
  for (int i = 0; i < 31; ++i) {
    int xx = (x >> (i + 1)) | (x & ((1 << i) - 1));
    if (xx != x) {
      int cur = DFS(xx);
      buf.push_back(cur);
    }
  }
  ret = Mex(buf);
  return ret;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    cnt.clear();
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
      Calc(a[i]);
    }
    int ans = 0;
    for (const auto &it : cnt) {
      int cur = DFS(it.second);
      ans ^= cur;
    }
    printf("%s\n", ans ? "Mojtaba" : "Arpa");
  }
  return 0;
}
