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 int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 100 + 15;
map<int, vector<int> > f;
map<vector<int>, int> g;
int n;
int GetMex(vector<int>& x) {
  if (x.empty()) return 0;
  sort(x.begin(), x.end());
  if (*x.begin() != 0) return 0;
  for (int i = 1; i < x.size(); ++i)
    if (x[i] - x[i - 1] > 1) return x[i - 1] + 1;
  return x.back() + 1;
}
int Dfs(vector<int> x) {
  if (x.empty()) return 0;
  if (g.count(x)) return g[x];
  int& ans = g[x] = 0;
  vector<int> vi;
  int mx = *x.rbegin();
  for (int i = 1; i <= mx; ++i) {
    vector<int> newx = x;
    for (int j = 0; j < x.size(); ++j)
      if (newx[j] >= i) newx[j] -= i;
    sort(newx.begin(), newx.end(), greater<int>());
    while (!newx.empty() && newx.back() == 0) newx.pop_back();
    reverse(newx.begin(), newx.end());
    vi.push_back(Dfs(newx));
  }
  return ans = GetMex(vi);
}
int main(int argc, char* argv[]) {
  n = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    for (int j = 2; j * j <= x; ++j) {
      if (x % j) continue;
      int cnt = 0;
      while (x % j == 0) {
        x /= j;
        ++cnt;
      }
      f[j].emplace_back(cnt);
    }
    if (x > 1) f[x].emplace_back(1);
  }
  int ans = 0;
  for (auto&& x : f) {
    vector<int> y = x.second;
    sort(y.begin(), y.end());
    y.assign(y.begin(), unique(y.begin(), y.end()));
    ans ^= Dfs(y);
  }
  if (ans)
    puts("Mojtaba");
  else
    puts("Arpa");
  return 0;
}
