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;
int i, j, m, n, p, k, pr[300005], a[300005], ans;
int mp[1200005], vis[1200005];
void Prime(int x) {
  int i;
  for (i = 2; i * i <= x; ++i)
    while (x % i == 0) pr[++pr[0]] = i, x /= i;
  if (x > 1) pr[++pr[0]] = x;
}
int dfs(int x) {
  if (!x) return 0;
  if (vis[x]) return mp[x];
  vis[x] = 1;
  int ans = 0, i, mex[45];
  for (i = 0; i < 45; ++i) mex[i] = 0;
  for (i = 1; x >= (1 << i); ++i) {
    int p = (x & ((1 << i) - 1)) | (x >> i);
    if (p & 1) --p;
    mex[dfs(p)] = 1;
  }
  for (i = 0; mex[i]; ++i)
    ;
  return mp[x] = i;
}
int Dfs(int x) {
  if (x < 1048576) return mp[x];
  int i, mex[45];
  for (i = 0; i < 45; ++i) mex[i] = 0;
  for (i = 1; x >= (1 << i); ++i) {
    int p = (x & ((1 << i) - 1)) | (x >> i);
    if (p & 1) --p;
    mex[Dfs(p)] = 1;
  }
  for (i = 0; mex[i]; ++i)
    ;
  return i;
}
int mex[105];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), Prime(a[i]);
  sort(pr + 1, pr + pr[0] + 1);
  pr[0] = unique(pr + 1, pr + pr[0] + 1) - (pr + 1);
  for (i = 0; i < (1 << 20); ++i) dfs(i);
  for (i = 1; i <= pr[0]; ++i) {
    int now = 0;
    for (j = 1; j <= n; ++j) {
      int s = 0;
      while (a[j] % pr[i] == 0) a[j] /= pr[i], ++s;
      now |= (1 << s);
    }
    if (now & 1) --now;
    ans ^= Dfs(now);
  }
  if (!ans)
    printf("Arpa");
  else
    printf("Mojtaba");
}
