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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int marked[1000000 / 64 + 2];
vector<int> primes;
bool isPrime(int x) {
  return (x > 1) &&
         ((x == 2) ||
          ((x & 1) && (!((marked[x >> 6] & (1 << ((x & 63) >> 1)))))));
}
void seive(int n) {
  int i, j;
  for (i = 3; i * i <= n; i += 2)
    if (!(marked[i >> 6] & (1 << ((i & 63) >> 1))))
      for (j = i * i; j <= n; j += i << 1)
        marked[j >> 6] |= (1 << ((j & 63) >> 1));
  primes.push_back(2);
  for (i = 3; i <= n; i += 2)
    if (isPrime(i)) primes.push_back(i);
}
map<int, int> mp, dp;
int f(int mask) {
  ;
  if (mask == 0) return 0;
  if (dp.find(mask) != dp.end()) return dp[mask];
  ;
  set<int> g;
  int i;
  for (i = 1; i < 32 - __builtin_clz(mask); i++) {
    g.insert(f(((mask >> (i + 1)) << 1) | (mask & ((1 << i) - 1))));
  }
  i = 0;
  for (int it : g) {
    if (it != i) break;
    i++;
    ;
  }
  return dp[mask] = i;
}
int main() {
  seive(100000);
  ;
  ;
  int n, i, j, a, cnt, nim = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    for (j = 0; j < primes.size() && primes[j] * primes[j] <= a; j++) {
      int v = primes[j];
      if (a % v == 0) {
        cnt = 0;
        while (a % v == 0) {
          cnt++;
          a /= v;
        }
        mp[v] |= (1 << (cnt));
      }
    }
    if (a > 1) {
      mp[a] |= (1 << 1);
    }
  }
  for (auto it : mp) {
    ;
    nim ^= f(it.second);
  }
  printf("%s\n", (nim) ? "Mojtaba" : "Arpa");
  return 0;
}
