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 long long MAXN = 100100;
const long long INF = 0x3f3f3f3f3f;
using namespace std;
bool isprime[MAXN];
long long prime[MAXN];
void linear() {
  long long cnt = -1;
  memset(isprime, 1, sizeof(isprime));
  isprime[1] = false;
  for (int i = 2; i <= MAXN; i++) {
    if (isprime[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= MAXN; j++) {
      isprime[i * prime[j]] = false;
      if (i % prime[j] == 0) break;
    }
  }
}
map<long long, long long> SG, num;
long long getSG(long long a) {
  if (a == 0) return 0;
  if (SG.count(a)) return SG[a];
  map<long long, long long> vis;
  for (long long m = 1; (a >> (m - 1)) > 0; ++m) {
    vis[getSG((a >> m) | (a & ((1 << m - 1) - 1)))] = 1;
  }
  for (long long i = 0; i <= a; ++i)
    if (!vis[i]) return SG[a] = i;
}
int main(int argc, char const *argv[]) {
  long long n;
  linear();
  while (cin >> n) {
    while (n--) {
      long long nn;
      cin >> nn;
      for (long long i = 0; prime[i] * prime[i] <= nn; ++i) {
        if (nn % prime[i]) continue;
        long long po;
        for (po = 0; !(nn % prime[i]); nn /= prime[i], po++)
          if (!num.count(prime[i])) num[prime[i]] = 0;
        num[prime[i]] |= 1 << (po - 1);
      }
      if (nn != 1) {
        if (!num.count(nn)) num[nn] = 0;
        num[nn] |= 1;
      }
    }
    map<long long, long long>::iterator it;
    long long ans = 0;
    long long te;
    for (it = num.begin(); it != num.end(); ++it) {
      te = (*it).second, ans ^= getSG(te);
    }
    if (ans)
      printf("Mojtaba\n");
    else
      printf("Arpa\n");
  }
  return 0;
}
