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 int MOD = 1e9 + 7;
const int N = 1e5 + 10;
const long long MAX = 1e18, OO = 1e16;
bool prime[N];
vector<int> pr;
map<int, int> me;
map<int, int> dp;
int n;
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int i = 2; i * i < N; ++i) {
    if (!prime[i]) continue;
    for (int j = i + i; j < N; j += i) {
      prime[j] = false;
    }
  }
  for (int i = 2; i < N; ++i) {
    if (prime[i]) pr.push_back(i);
  }
}
int MEX(set<int> &s) {
  int cnt = 0;
  while (s.count(cnt)) ++cnt;
  return cnt;
}
int solve(bitset<31> mask) {
  int org = mask.to_ullong();
  if (org == 0) return 0;
  if (dp.find(org) != dp.end()) {
    return dp[org];
  }
  set<int> st;
  int idx = -1;
  for (int i = 30; i >= 0; --i) {
    if (mask[i]) {
      idx = i;
      break;
    }
  }
  for (int shift = 0; shift <= idx; ++shift) {
    bitset<31> bit = mask;
    for (int after = shift; after <= idx; ++after) {
      if (bit[after]) {
        bit[after] = 0;
        if (after - shift - 1 >= 0) bit[after - shift - 1] = 1;
      }
    }
    st.insert(solve(bit));
  }
  return dp[mask.to_ullong()] = MEX(st);
}
int main() {
  sieve();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    for (int j = 0; j < pr.size(); ++j) {
      int cnt = 0;
      while (x % pr[j] == 0) {
        x /= pr[j];
        ++cnt;
      }
      if (cnt >= 1) me[pr[j]] |= (1 << cnt - 1);
    }
    if (x > 1) me[x] |= (1 << 0);
  }
  int sofar = 0;
  for (auto c : me) {
    sofar ^= solve(c.second);
  }
  if (sofar == 0)
    cout << "Arpa" << endl;
  else
    cout << "Mojtaba" << endl;
}
