Codeforces Round 432 (Div. 1, based on IndiaHacks Final Round 2017)
C
Arpa and a game with Mojtaba
Codeforces
2200
原题
← BD →
翻译中…

Mojtaba and Arpa are playing a game. They have a list of $*n*$ numbers in the game.

翻译中…

In a player's turn, he chooses a number $*p*^{*k*}$ (where $*p*$ is a prime number and $*k*$ is a positive integer) such that $*p*^{*k*}$ divides at least one number in the list. For each number in the list divisible by $*p*^{*k*}$, 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.

翻译中…

Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.

翻译中…

Input

翻译中…

The first line contains a single integer $*n*$ ($1 ≤ *n* ≤ 100$) — the number of elements in the list.

翻译中…

The second line contains $*n*$ integers $*a*_{1}, *a*_{2}, ..., *a*_{*n*}$ ($1 ≤ *a*_{*i*} ≤ 10^{9}$) — the elements of the list.

翻译中…

Output

翻译中…

If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).

翻译中…

You can print each letter in any case (upper or lower).

翻译中…

Note

翻译中…

In the first sample test, Mojtaba can't move.

翻译中…

In the second sample test, Mojtaba chooses $*p* = 17$ and $*k* = 1$, then the list changes to $[1, 1, 1, 1]$.

翻译中…

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.

翻译中…

样例

Input #1
4
1 1 1 1
Output #1
Arpa
InputCopy OutputCopy
Input #2
4
1 1 17 17
Output #2
Mojtaba
InputCopy OutputCopy
Input #3
4
1 1 17 289
Output #3
Arpa
InputCopy OutputCopy
Input #4
5
1 2 3 4 5
Output #4
Arpa
InputCopy OutputCopy
题目信息
题号
850C
时间限制
1 s
内存限制
256 MB
标签
位运算
动态规划
博弈论

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