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 = 2100000;
int sg[N];
int mex[40];
int ans;
int vis[20][40];
int ck, kk[100010];
int a[200];
void decompose(int x, int id) {
  int i;
  for (i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ck++;
      kk[ck] = i;
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) {
    ck++;
    kk[ck] = x;
  }
}
int dfs(int i, int cur) {
  int j, x;
  if (i < (1 << 20)) return sg[i];
  for (j = 0; j <= 32; j++) vis[cur][j] = 0;
  for (j = 1; j <= 30; j++) {
    if (!(i >> (j - 1))) break;
    x = (i >> j) | (i & ((1 << j - 1) - 1));
    vis[cur][dfs(x, cur + 1)] = 1;
  }
  for (j = 0; j <= 32; j++)
    if (!vis[cur][j]) {
      return j;
    }
}
int main() {
  int i, j, x, n;
  int now, tmp;
  sg[0] = 0;
  for (i = 1; i <= (1 << 20) - 1; i++) {
    for (j = 0; j <= 22; j++) mex[j] = 0;
    for (j = 1; j <= 20; j++) {
      if (i >> (j - 1)) {
        x = (i >> j) | (i & ((1 << j - 1) - 1));
        mex[sg[x]] = 1;
      }
    }
    for (j = 0; j <= 22; j++)
      if (!mex[j]) {
        sg[i] = j;
        break;
      }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++) {
    decompose(a[i], i);
  }
  sort(kk + 1, kk + 1 + ck);
  ck = unique(kk + 1, kk + 1 + ck) - kk - 1;
  for (i = 1; i <= ck; i++) {
    tmp = 0;
    for (j = 1; j <= n; j++) {
      now = 0;
      x = a[j];
      while (x % kk[i] == 0) {
        x /= kk[i];
        now++;
      }
      if (now) {
        tmp |= (1 << now - 1);
      }
    }
    ans ^= dfs(tmp, 0);
  }
  if (ans) {
    printf("Mojtaba\n");
  } else
    printf("Arpa\n");
}
