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 C = 26;
void add(int& x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e2 + 10;
int n, a[MAXN];
map<set<int>, int> memo;
int mex(const set<int>& s) {
  int ans = 0;
  while (s.count(ans)) ans++;
  return ans;
}
int go(set<int>& s) {
  if (s.empty()) return 0;
  if (memo.count(s)) return memo[s];
  int mx = *s.rbegin();
  set<int> grundy;
  for (int take = 1; take <= mx; take++) {
    set<int> ns;
    for (int x : s)
      if (x > take)
        ns.insert(x - take);
      else if (x < take)
        ns.insert(x);
    grundy.insert(go(ns));
  }
  return memo[s] = mex(grundy);
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  map<int, set<int>> games;
  for (int i = 0; i < n; i++) {
    int N = a[i];
    for (int j = 2; j * j <= N; j++) {
      int cnt = 0;
      while (N % j == 0) {
        ++cnt;
        N /= j;
      }
      if (cnt) games[j].insert(cnt);
    }
    if (N > 1) games[N].insert(1);
  }
  int s = 0;
  for (auto& p : games) {
    memo.clear();
    s ^= go(p.second);
  }
  if (s)
    cout << "Mojtaba\n";
  else
    cout << "Arpa\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}
