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 = 100005;
int n, cnt, m, ans;
int pri[N], p[N], a[N];
bool vis[N];
map<int, int> mp1, mp2;
void init(int n) {
  int i, j;
  for (i = 2; i <= n; i++) {
    if (!vis[i]) pri[++cnt] = i;
    for (j = 1; j <= cnt && pri[j] * i <= n; j++) {
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
void split(int w) {
  int i, j;
  for (i = 1; i <= cnt && pri[i] <= w; i++)
    if (w % pri[i] == 0) {
      p[++m] = pri[i];
      j = 0;
      while (w % pri[i] == 0) {
        w /= pri[i];
        j++;
      }
      mp1[pri[i]] |= (1 << j - 1);
    }
  if (w > 1) {
    p[++m] = w;
    mp1[w] |= 1;
  }
}
int dfs(int s) {
  if (!s) return 0;
  if (mp2.count(s)) return mp2[s];
  int i;
  bool vis[32];
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= 30; i++) {
    if (!(s >> i - 1)) break;
    vis[dfs((s >> i) | (s & ((1 << i - 1) - 1)))] = 1;
  }
  for (i = 0; i <= 30; i++)
    if (!vis[i]) {
      mp2[s] = i;
      break;
    }
  return mp2[s];
}
int main() {
  int i;
  init(100000);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    split(a[i]);
  }
  sort(p + 1, p + 1 + m);
  m = unique(p + 1, p + 1 + m) - p - 1;
  for (i = 1; i <= m; i++) ans ^= dfs(mp1[p[i]]);
  if (ans)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
  return 0;
}
