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;
bool isp[40000];
vector<int> pl;
map<int, int> nim, gr;
int foo(int st) {
  if (gr.find(st) != gr.end()) return gr[st];
  map<int, char> v;
  for (int i = 1; i < 31; i++) {
    if ((1 << (i - 1)) > st) break;
    int s = st >> i;
    s |= st & ((1 << (i - 1)) - 1);
    v[foo(s)] = true;
  }
  int l = -1;
  bool ok = false;
  for (auto p : v) {
    if (p.first > l + 1) {
      ok = true;
      gr[st] = l + 1;
      break;
    } else
      l = p.first;
  }
  if (!ok) gr[st] = l + 1;
  return gr[st];
}
int main() {
  isp[2] = true;
  for (int i = 3; i < 40000; i += 2) isp[i] = true;
  for (int i = 3; i * i < 40000; i += 2) {
    if (!isp[i]) continue;
    for (int j = i * i; j < 40000; j += i) isp[j] = false;
  }
  pl.push_back(2);
  for (int i = 3; i < 40000; i += 2)
    if (isp[i]) pl.push_back(i);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    for (int p : pl) {
      if (p > x) break;
      if (x % p) continue;
      int c = 0;
      while (x % p == 0) {
        c++;
        x /= p;
      }
      nim[p] |= 1 << (c - 1);
    }
    if (x > 1) nim[x] |= 1;
  }
  int x = 0;
  gr[0] = 0;
  for (auto p : nim) {
    x ^= foo(p.second);
  }
  puts(x ? "Mojtaba" : "Arpa");
}
