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 int MAXN = 5e4 + 5;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
template <class T>
inline bool read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
const int MAXM = 40000;
int prime[MAXM + 10], tot = 0;
bool vis[MAXM + 10];
inline void f() {
  for (register int i = 2; i <= MAXM; i++) {
    if (!vis[i]) prime[tot++] = i;
    for (register int j = 0; j < tot && i * prime[j] <= MAXM; j++)
      vis[i * prime[j]] = 1;
  }
}
int a[105];
map<int, int> SG, num;
int getSG(int x) {
  if (x == 0) return 0;
  if (SG.count(x)) return SG[x];
  map<int, int> vis;
  vis.clear();
  int p = x, t = 0;
  while (p) {
    p /= 2;
    t++;
  }
  for (int i = 1; i <= t; ++i)
    vis[getSG(x >> i | (x & ((1 << i - 1) - 1)))] = 1;
  for (int i = 0;; ++i) {
    if (!vis.count(i)) return SG[x] = i;
  }
}
int main() {
  f();
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    int x = a[i];
    for (int j = 0; j < tot; ++j) {
      int p = prime[j];
      if (p * p > a[i]) break;
      int t = 0;
      while (x % p == 0) {
        x /= p;
        t++;
      }
      if (t) num[p] = num[p] | 1 << (t - 1);
    }
    if (x != 1) num[x] = num[x] | 1;
  }
  map<int, int>::iterator it;
  int ans = 0;
  for (it = num.begin(); it != num.end(); ++it) ans ^= getSG(it->second);
  if (ans)
    cout << "Mojtaba";
  else
    cout << "Arpa";
  return 0;
}
