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 N = 10005;
struct myhash {
  static const int mod = 20000003;
  int fir[mod], nxt[mod / 5], pb[mod / 5], tp[mod / 5], tot;
  void ins(int x, int y) {
    int w = x % mod;
    tp[++tot] = y, pb[tot] = x;
    nxt[tot] = fir[w], fir[w] = tot;
  }
  int que(int x) {
    int w = x % mod;
    for (int i = fir[w]; i; i = nxt[i])
      if (pb[i] == x) return tp[i];
    return -1;
  }
} sg;
struct node {
  int a, b;
  bool operator<(const node& t) const { return a < t.a; }
} p[N];
int Sg(int x) {
  int tmp = sg.que(x);
  if (~tmp) return tmp;
  int i;
  bool vis[33];
  memset(vis, 0, sizeof(vis));
  for (i = 1; x >> i; i++) vis[Sg(x >> i | x & (1 << i) - 1)] = 1;
  for (i = 0; vis[i]; i++)
    ;
  sg.ins(x, i);
  return i;
}
int n, tot;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = (1), i_end = (n); i <= i_end; i++) {
    int x, q;
    cin >> x;
    for (int i = 2; i * i <= x && x ^ 1; i++)
      if (x % i == 0) {
        int a = i, b = 0;
        while (x % i == 0) b++, x /= i;
        p[++tot] = (node){a, b};
      }
    if (x ^ 1) p[++tot] = (node){x, 1};
  }
  sort(p + 1, p + tot + 1);
  int ans = 0;
  n = tot;
  for (int i = 1, j; i <= n; i = j) {
    int now = 1 << p[i].b;
    for (j = i + 1; j <= n && p[j].a == p[i].a; j++) {
      now |= 1 << p[j].b;
    }
    ans ^= Sg(now | 1);
  }
  puts((ans > 0) ? "Mojtaba" : "Arpa");
}
