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 N, A[110];
const int up = 1e9 + 10;
unordered_map<int, vector<int> > t;
unordered_map<long long, int> sg;
const long long MOD = 1e9 + 7, seed = 233;
bool vis[40];
void ins(int v, int cnt) { t[v].push_back(cnt); }
long long calc(vector<int> v) {
  if (v.empty()) return 0;
  sort(v.begin(), v.end());
  long long hsh = 0;
  for (auto tt : v) hsh = (hsh * seed + tt) % MOD;
  if (sg.find(hsh) != sg.end()) return sg[hsh];
  vector<int> tmp, mex;
  mex.clear();
  for (int tt = 1; tt <= v.back(); tt++) {
    tmp.clear();
    for (auto it : v) {
      if (it < tt) tmp.push_back(it);
      if (it > tt) tmp.push_back(it - tt);
    }
    mex.push_back(calc(tmp));
  }
  memset(vis, false, sizeof(vis));
  for (auto tt : mex) vis[tt] = true;
  for (int i = 0; i <= 10000; i++)
    if (!vis[i]) return sg[hsh] = i;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= N; i++) {
    int tmp = A[i];
    for (int j = 2; j * j <= up; j++)
      if (tmp % j == 0) {
        int cnt = 0;
        while (tmp % j == 0) {
          cnt++;
          tmp /= j;
        }
        ins(j, cnt);
      }
    if (tmp > 1) ins(tmp, 1);
  }
  int res = 0;
  for (auto it : t) {
    res ^= calc(it.second);
  }
  puts(res ? "Mojtaba" : "Arpa");
}
