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>
template <class T>
inline void read(T &x) {
  int t;
  while (((t = getchar()) < '0' || t > '9') && t != '-')
    ;
  bool flag = false;
  if (t == '-') flag = true, t = getchar();
  x = t - '0';
  while ((t = getchar()) >= '0' && t <= '9') x = x * 10 + (t - '0');
  if (flag) x = -x;
}
using namespace std;
const int maxn = 110;
int n, N;
pair<int, int> d[maxn * 100];
map<int, int> sg;
int getsg(int w) {
  if (sg.find(w) != sg.end()) return sg[w];
  int cur = 0;
  for (int k = 0, to; w >> k; ++k) {
    to = (w >> (k + 1)) | (w & ((1 << k) - 1));
    cur |= (1 << getsg(to));
  }
  int res = 0;
  while (cur & (1 << res)) ++res;
  return sg[w] = res;
}
int main() {
  read(n);
  for (int i = 1, ori, tmp; i <= n; ++i) {
    read(ori), tmp = ori;
    for (int j = 2; j * j <= ori; ++j) {
      int cnt = 0;
      while (tmp % j == 0) tmp /= j, ++cnt;
      if (cnt) d[++N] = pair<int, int>(j, cnt);
    }
    if (tmp != 1) d[++N] = pair<int, int>(tmp, 1);
  }
  sort(d + 1, d + N + 1);
  sg[0] = 0;
  int sg_all = 0;
  for (int l = 1, r = 1; l <= N; l = ++r) {
    while (r < N && d[r].first == d[r + 1].first) ++r;
    int val = 0;
    for (int i = l; i <= r; ++i) {
      val |= (1 << (d[i].second - 1));
    }
    sg_all ^= getsg(val);
  }
  if (sg_all)
    puts("Mojtaba");
  else
    puts("Arpa");
  return 0;
}
