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 INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e5 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = a * a % M)
    if (b & 1) ret = 1ll * ret * a % M;
  return ret;
}
map<vector<int>, int> HASH;
int calc(vector<int> now) {
  if (HASH.count(now)) return HASH[now];
  set<int> mex;
  int MX = 0, i;
  for (int val : now) MX = max(MX, val);
  for (i = 1; i <= MX; i++) {
    vector<int> down;
    for (int val : now) {
      if (val == i) continue;
      if (val > i)
        down.push_back(val - i);
      else
        down.push_back(val);
    }
    sort(down.begin(), down.end());
    down.erase(unique(down.begin(), down.end()), down.end());
    mex.insert(calc(down));
  }
  int ret = 0;
  while (mex.count(ret)) ret++;
  return HASH[now] = ret;
}
int cnt;
int n, k;
int i, j;
int ans;
vector<int> val[maxn];
map<int, int> MP;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &k);
    for (j = 2; j * j <= k; j++) {
      int tot = 0;
      while (k % j == 0) k /= j, tot++;
      if (tot) {
        if (!MP.count(j)) MP[j] = ++cnt;
        val[MP[j]].push_back(tot);
      }
    }
    if (k != 1) {
      if (!MP.count(k)) MP[k] = ++cnt;
      val[MP[k]].push_back(1);
    }
  }
  for (i = 1; i <= cnt; i++)
    sort(val[i].begin(), val[i].end()),
        val[i].erase(unique(val[i].begin(), val[i].end()), val[i].end());
  for (i = 1; i <= cnt; i++)
    if (val[i].size()) ans ^= calc(val[i]);
  if (ans)
    puts("Mojtaba");
  else
    puts("Arpa");
}
