Codeforces451A-Game With Sticks-思维

After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of n horizontal and m vertical sticks.

An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.

In the grid shown below, n = 3 and m = 3. There are n + m = 6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are n·m = 9 intersection points, numbered from 1 to 9.

The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).

Assume that both players play optimally. Who will win the game?

Input

The first line of input contains two space-separated integers, n and m (1 ≤ n, m ≤ 100).

Output

Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.

Examples

input

Copy

2 2

output

Copy

Malvika

input

Copy

2 3

output

Copy

Malvika

input

Copy

3 3

output

Copy

Akshat

Note

Explanation of the first sample:

The grid has four intersection points, numbered from 1 to 4.

If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.

Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.

In the empty grid, Akshat cannot make any move, hence he will lose.

Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.

题意：

给出n和m，表示有n+m根棍子，n*m个交叉点，每次每个人可以选择一个点，拿走涉及到该点的所有棍子。A先拿，谁先拿完全部棍子谁赢，两个人都是最优策略。

问最终谁赢。

思路：

取n和m的最小值，再判断一下奇偶性即可。

发现好像很多乍一眼看上去像博弈的题，其实都是代码很短的思维题。。

 #include<bits/stdc++.h>
 using namespace std;

 int main()
 {
     int n,m;
     while(~scanf("%d %d",&n,&m))
     {
       int minn=min(n,m);
       if(minn%==)
         printf("Malvika\n");
       else
         printf("Akshat\n");
     }
     return ;
 }