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 = 1000000 + 5;
int n, tot, ans, pri[Maxn];
bool vis[Maxn];
map<int, int> st, sg;
void w(int x) {
  for (int i = 1; pri[i] * pri[i] <= x; i++) {
    int t = 0;
    while (x % pri[i] == 0) {
      x /= pri[i];
      t++;
    }
    if (t) st[pri[i]] |= 1 << (t - 1);
  }
  if (x != 1) st[x] |= 1;
}
int ssg(int x) {
  int p = x, t = 0;
  map<int, int> vis;
  if (x == 0) return 0;
  if (sg.count(x)) return sg[x];
  while (p) {
    p >>= 1;
    t++;
  }
  for (int i = 1; i <= t; i++)
    vis[ssg((x >> i) | (x & (1 << (i - 1)) - 1))] = 1;
  for (int i = 0;; i++)
    if (!vis[i]) return sg[x] = i;
}
int main() {
  for (int i = 2; i <= 1000000; i++) {
    if (vis[i]) continue;
    pri[++tot] = i;
    for (int j = 1; j * i <= 1000000; j++) vis[i * j] = 1;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    w(x);
  }
  map<int, int>::iterator it;
  for (it = st.begin(); it != st.end(); it++) ans = ans ^ ssg(it->second);
  if (ans)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
  return 0;
}
