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 N = 101010;
int n;
int a[N];
long long sta[N];
bool vis[N];
vector<int> p;
map<long long, int> sg;
map<int, int> Vis;
void init() {
  for (int i = (2); i < (N); i++)
    if (!vis[i]) {
      p.push_back(i);
      for (int j = i + i; j < N; j += i) vis[j] = 1;
    }
}
int solve(long long s) {
  if (sg.count(s)) return sg[s];
  bool mex[33] = {0};
  for (int i = 1;; ++i) {
    mex[solve((s >> i) | (s & (1 << i - 1) - 1))] = 1;
    if ((s >> i) == 0) break;
  }
  for (int i = 0;; ++i)
    if (!mex[i]) return sg[s] = i;
}
int main() {
  init();
  while (~scanf("%d", &n)) {
    memset(sta, 0, sizeof(sta));
    sg.clear();
    Vis.clear();
    for (int i = (1); i < (n + 1); i++) scanf("%d", a + i);
    int ans = 0;
    for (int i = (1); i < (n + 1); i++) {
      for (int j = (0); j < ((int)p.size()); j++) {
        int cnt = 0;
        while (a[i] % p[j] == 0) a[i] /= p[j], ++cnt;
        if (cnt) sta[j] |= 1ll << cnt - 1;
      }
      if (a[i] != 1 && !Vis[a[i]]) Vis[a[i]] = 1, ans ^= 1;
    }
    sg[0] = 0;
    for (int j = (0); j < ((int)p.size() + 1); j++)
      if (sta[j]) ans ^= solve(sta[j]);
    printf("%s\n", ans ? "Mojtaba" : "Arpa");
  }
  return 0;
}
