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;
int shortenAndGetPower(int divider, int &x) {
  int power = 0;
  while (0 == x % divider) {
    x /= divider;
    ++power;
  }
  return power;
}
list<pair<int, int> > factorize(int x) {
  list<pair<int, int> > factorization;
  if (0 == x % 2) {
    factorization.emplace_back(2, shortenAndGetPower(2, x));
  }
  for (int divider = 3; 1LL * divider * divider <= x; divider += 2) {
    if (0 == x % divider) {
      factorization.emplace_back(divider, shortenAndGetPower(divider, x));
    }
  }
  if (1 < x) {
    factorization.emplace_back(x, 1);
  }
  return factorization;
}
map<vector<int>, long long> spragueGrundyMemory;
long long calculateSpragueGrundy(vector<int> powers) {
  sort(powers.begin(), powers.end());
  if (spragueGrundyMemory.count(powers)) {
    return spragueGrundyMemory[powers];
  }
  unsigned long long spragueGrundyUsed = 0;
  for (int power = 1; power <= powers.back(); ++power) {
    vector<int> newPowers;
    for (int i = 0; i < (int)powers.size(); ++i) {
      if (powers[i] != power) {
        newPowers.push_back(powers[i] < power ? powers[i] : powers[i] - power);
      }
    }
    spragueGrundyUsed |= 1LL << calculateSpragueGrundy(newPowers);
  }
  long long mex;
  for (mex = 0; (spragueGrundyUsed >> mex) & 1; ++mex)
    ;
  return spragueGrundyMemory[powers] = mex;
}
bool isMojtaba(int n, int a[]) {
  map<int, vector<int> > primeToPowers;
  for (int i = 0; i < n; ++i) {
    const auto primePowers = factorize(a[i]);
    for (const auto &primePower : primePowers) {
      primeToPowers[primePower.first].push_back(primePower.second);
    }
  }
  spragueGrundyMemory.clear();
  spragueGrundyMemory[vector<int>()] = 0;
  long long spragueGrundy = 0;
  for (const auto &primePowers : primeToPowers) {
    spragueGrundy ^= calculateSpragueGrundy(primePowers.second);
  }
  return spragueGrundy > 0;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    int a[n];
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    if (isMojtaba(n, a)) {
      printf("Mojtaba\n");
    } else {
      printf("Arpa\n");
    }
  }
  return 0;
}
