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>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
const int NMAX = 105;
int n;
int a[NMAX];
vector<pair<int, int> > v;
map<int, int> g;
int grundy(int mask) {
  if (g.count(mask)) return g[mask];
  if (mask == 0) return 0;
  int nmask;
  vector<int> v;
  for (int i = 1; i <= int(30); i++) {
    nmask = (mask >> i) | (mask & ((1 << (i - 1)) - 1));
    if (nmask != mask)
      v.push_back(grundy(nmask));
    else
      break;
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0;; i++)
    if ((int)v.size() <= i || i < v[i]) return g[mask] = i;
}
int main() {
  cin >> n;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  for (int i = 0; i < int(n); i++) {
    int x = a[i];
    for (int j = 2; j * j <= x; j++) {
      if (x % j == 0) {
        int cnt = 0;
        while (x % j == 0) {
          x /= j;
          cnt++;
        }
        v.push_back(make_pair(j, cnt));
      }
    }
    if (x > 1) v.push_back(make_pair(x, 1));
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  int l = 0;
  int m = (int)v.size();
  int ans = 0;
  while (l < m) {
    int r = l;
    int mask = 0;
    while (r < m && v[r].first == v[l].first) {
      mask |= (1 << (v[r].second - 1));
      r++;
    }
    ans ^= grundy(mask);
    l = r;
  }
  if (ans)
    cout << "Mojtaba\n";
  else
    cout << "Arpa\n";
  return 0;
}
