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 double sn = 1e-6;
int n;
map<int, long long> mpp;
map<long long, int> dp;
map<long long, bool> vis;
int mex(set<int> &s) {
  int c = 0;
  while (s.find(c) != s.end()) c++;
  return c;
}
int grundy(long long v) {
  if (v == 0) return 0;
  if (vis[v]) return dp[v];
  set<int> s;
  for (int i = 1; (1 << (i - 1)) <= v; i++) {
    long long v2 = 0;
    if (i != 1) v2 = ((1 << (i - 1)) - 1) & v;
    s.insert(grundy((v >> i) | v2));
  }
  vis[v] = true;
  return dp[v] = mex(s);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int v;
    scanf("%d", &v);
    for (long long j = 2; j * j <= v; j++) {
      int c = 0;
      while (v % j == 0) {
        v /= j;
        c++;
      }
      if (c != 0) {
        mpp[j] |= (1 << (c - 1));
      }
    }
    if (v != 1) mpp[v] |= 1;
  }
  int xs = 0;
  for (auto v : mpp) {
    xs ^= grundy(v.second);
  }
  if (xs == 0)
    printf("Arpa\n");
  else
    printf("Mojtaba\n");
  return 0;
}
