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 long long mod = 1e9 + 7;
long double PI = acosl(-1);
const int INF = 1e8 + 1;
const int nmax = 2e5 + 5;
const int MAXLG = log2(nmax) + 1;
int primemask[nmax];
unordered_map<int, bool> mp;
unordered_map<int, int> dp;
int getmex(int mask) {
  if (dp.find(mask) != dp.end()) return dp[mask];
  unordered_set<int> mex;
  for (int i = 0; i < 31; i++) {
    int x = (mask >> (i + 1)) | (mask & ((1 << i) - 1));
    if (x != mask) {
      int y = getmex(x);
      mex.insert(y);
    }
  }
  for (int i = 0;; i++)
    if (mex.find(i) == mex.end()) return dp[mask] = i;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  dp[0] = 0;
  int n;
  cin >> n;
  int boroprime = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        int cnt = 0;
        while (x % i == 0) cnt++, x /= i;
        primemask[i] |= (1 << (cnt - 1));
      }
    }
    if (x > nmax) {
      if (!mp[x]) mp[x] = true, boroprime++;
    } else {
      primemask[x] |= 1;
    }
  }
  boroprime %= 2;
  vector<int> msks;
  for (int i = 2; i < nmax; i++) {
    if (primemask[i]) {
      msks.push_back(primemask[i]);
    }
  }
  int ans = 0;
  ans ^= boroprime;
  for (int i = 0; i < msks.size(); i++) ans ^= getmex(msks[i]);
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
}
