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 n, cnt, ans;
int a[100010], p[100010], num[100010], sum[100010], dy[100010];
bool no[100010];
map<int, int> st, sg;
void shai() {
  int t, j;
  no[1] = true;
  for (int i = 2; i < 100010; i++) {
    if (!no[i]) p[++p[0]] = i, dy[i] = i;
    t = p[1] * i;
    j = 1;
    while (j <= p[0] && t < 100010) {
      no[t] = true;
      if (i % p[j] == 0) {
        dy[i] = p[j];
        break;
      }
      t = p[++j] * i;
    }
  }
}
void prework() {
  cnt = p[0];
  st.clear();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = 1; j <= p[0]; j++) {
      int d = 0;
      while (a[i] % p[j] == 0 && a[i] > 1) a[i] /= p[j], d++;
      if (d > 0) st[p[j]] |= 1 << (d - 1);
    }
    if (a[i] > 1) st[a[i]] |= 1;
  }
}
int getsg(int x) {
  if (x == 0) return 0;
  if (sg.count(x)) return sg[x];
  map<int, int> vis;
  int p = x, t = 0;
  while (p) p >>= 1, t++;
  for (int i = 1; i <= t; i++)
    vis[getsg((x >> i) | (x & ((1 << (i - 1)) - 1)))] = 1;
  for (int i = 0; i < 100010; i++)
    if (!vis[i]) return sg[x] = i;
}
void mainwork() {
  ans = 0;
  map<int, int>::iterator it;
  for (it = st.begin(); it != st.end(); it++) ans = ans ^ getsg((*it).second);
}
void print() {
  if (ans)
    printf("Mojtaba\n");
  else
    printf("Arpa\n");
}
int main() {
  shai();
  while (~scanf("%d", &n)) {
    prework();
    mainwork();
    print();
  }
  return 0;
}
