Problem: Mike and Ann are sitting in the classroom. The lesson is boring, so they decided to play an interesting game. Fortunately, all they need to play this game is a string s and a number k (0 ≤ k < |s|).

At the beginning of the game, players are given a substring of s with left border l and right border r, both equal to k (i.e. initially l=r=k). Then players start to make moves one by one, according to the following rules:

  * A player chooses l^{\prime} and r^{\prime} so that l^{\prime} ≤ l, r^{\prime} ≥ r and s[l^{\prime}, r^{\prime}] is lexicographically less than s[l, r]. Then the player changes l and r in this way: l := l^{\prime}, r := r^{\prime}.
  * Ann moves first.
  * The player, that can't make a move loses.



Recall that a substring s[l, r] (l ≤ r) of a string s is a continuous segment of letters from s that starts at position l and ends at position r. For example, "ehn" is a substring (s[3, 5]) of "aaaehnsvz" and "ahz" is not.

Mike and Ann were playing so enthusiastically that they did not notice the teacher approached them. Surprisingly, the teacher didn't scold them, instead of that he said, that he can figure out the winner of the game before it starts, even if he knows only s and k.

Unfortunately, Mike and Ann are not so keen in the game theory, so they ask you to write a program, that takes s and determines the winner for all possible k.

Input

The first line of the input contains a single string s (1 ≤ |s| ≤ 5 ⋅ 10^5) consisting of lowercase English letters.

Output

Print |s| lines.

In the line i write the name of the winner (print Mike or Ann) in the game with string s and k = i, if both play optimally

Examples

Input


abba


Output


Mike
Ann
Ann
Mike


Input


cba


Output


Mike
Mike
Mike

Solution (in JAVA):

Solution:
//package com.company;

//import java.io.BufferedReader;
//import java.io.IOException;
//import java.io.IOException;
//import java.io.InputStreamReader;
//import java.math.BigInteger;
//import java.nio.file.Paths;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        // write your code here
        // BufferedReader s=new BufferedReader(new InputStreamReader(System.in));
        //BufferedReader s=new BufferedReader(new InputStreamReader(System.in));    
        StringBuilder sb = new StringBuilder();
        Scanner s=new Scanner(System.in);
        //     String[] str=s.readLine().trim().split("\\s+");
     //   Integer n=s.nextInt();
        char[] a=s.next().toCharArray();int[] c=new int[123];
        for(int i=0;i<a.length;i++){
            c[a[i]]++;int flag1=0;
            for(int j=97;j<a[i];j++){
                if(c[j]>0){
                    flag1=1;break;
                }
            }if(flag1==1){
               sb.append("Ann\n");
            }else
                sb.append("Mike\n");
        }System.out.println(sb.toString());
    }

    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a*b)/gcd(a, b);
    }
    static int[] vis;
    static int count;
    static int[] flag;
    static int another;
    static int connected;static int huf=0;
    static void dfs(int i,LinkedList<Integer> [] adj){
        connected++;
        vis[i]=1;if(flag[i]==1) count++;
        if(adj[i]!=null){
            for(int j:adj[i]){
                if(vis[j]==0){
                    if(flag[i]==0) flag[j]=1; else flag[j]=0;

                    dfs(j,adj);
                } else{
                    if(flag[i]==flag[j]){
                        huf=1;
                    }
                }


            }}else{another++;}//return count;
    }






    static  Boolean isCyclicUtil(int v, Boolean visited[], int parent,LinkedList[] adj)
    {
        // Mark the current node as visited
        visited[v] = true;
        Integer i;

        // Recur for all the vertices adjacent to this vertex
        if(adj[v]!=null){
            Iterator<Integer> it = adj[v].iterator();
            while (it.hasNext())
            {
                i = it.next();

                // If an adjacent is not visited, then recur for that
                // adjacent
                if (!visited[i])
                {
                    if (isCyclicUtil(i, visited, v,adj))
                        return true;
                }

                // If an adjacent is visited and not parent of current
                // vertex, then there is a cycle.
                else if (i != parent)
                    return true;
            }}
        return false;
    }

    // Returns true if the graph contains a cycle, else false.
    static Boolean isCyclic(int V,LinkedList<Integer>[] adj)
    {
        // Mark all the vertices as not visited and not part of
        // recursion stack
        Boolean visited[] = new Boolean[V];
        for (int i = 0; i < V; i++)
            visited[i] = false;

        // Call the recursive helper function to detect cycle in
        // different DFS trees
        for (int u = 0; u < V; u++)
            if (!visited[u]) // Don't recur for u if already visited
                if (isCyclicUtil(u, visited, -1,adj))
                    return true;

        return false;
    }

}

class Student
{
    int rollno;
    int color;

    // Constructor
    public Student(int rollno, int color)
    {
        this.rollno = rollno;

        this.color=color;}
    // Used to print student details in main()
    public String toString()
    {
        return this.rollno + " " ;
    }
}

class Sortbyroll implements Comparator<Student>
{
    // Used for sorting in ascending order of
    // roll number
    public int compare(Student a, Student b)
    {
        return a.color - b.color;
    }
}