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;
long long pow(long long a, long long b, long long M) {
  long long t = 1;
  for (a %= M; b; b >>= 1) {
    if (b & 1) t = t * a % M;
    a = a * a % M;
  }
  return t;
}
void R(int &first) {
  static char s[22];
  scanf("%s", s);
  int f = 1;
  char *c = s;
  first = 0;
  if (*c == '-') f = -1, c++;
  for (; *c; c++) first = first * 10 + *c - '0';
  first *= f;
}
const int N = 105, M = 30;
int v[M], n, cnt;
map<int, int> q;
int dfs() {
  int j = 0;
  for (int i = M - 1; i >= int(1); i--)
    if (v[i]) {
      j = i;
      break;
    }
  int t = 0;
  for (int i = 1; i <= int(M); i++) t += v[i] << i;
  if (q.count(t)) return q[t];
  cnt++;
  if (!j) return 0;
  int g[N] = {0};
  for (int i = 1; i <= int(j); i++) {
    int t[M];
    for (int i = 0; i <= int(M - 1); i++) t[i] = v[i];
    for (int k = i; k <= int(M - 1); k++) v[k - i] |= v[k], v[k] = 0;
    g[dfs()] = 1;
    for (int i = 0; i <= int(M - 1); i++) v[i] = t[i];
  }
  for (int i = 0; i <= int(N - 1); i++)
    if (!g[i]) return q[t] = i;
  return q[t] = 0;
}
map<int, vector<int>> a;
int main() {
  cin >> n;
  for (int k = 1; k <= int(n); k++) {
    int first;
    R(first);
    for (int i = 2; i * i <= first; i++)
      if (first % i == 0) {
        int cnt = 0;
        while (first % i == 0) first /= i, cnt++;
        a[i].push_back(cnt);
      }
    if (first > 1) a[first].push_back(1);
  }
  int ans = 0;
  for (auto i : a) {
    memset(v, 0, sizeof(v));
    for (auto j : i.second) {
      v[j] = 1;
    }
    ans ^= dfs();
  }
  if (ans)
    puts("Mojtaba");
  else
    puts("Arpa");
}
