/*
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11800    Accepted Submission(s): 3673
Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

Input
One positive integer on each line, the value of n.

Output
If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.

Sample Input
2
5
 

Sample Output
2^? mod 2 = 1
2^4 mod 5 = 1

*/
#include<stdio.h>
int main(){
 int i,j,n,s;
 while(~scanf("%d",&n)){
  if(n==1||n%2==0)  printf("2^? mod %d = 1\n",n);
  else{
   __int64  s=1;
   int i;
   for(i=1;;i++){
    s*=2;
    s%=n;        //控制s的值在一定范围内，乘2的同时对n取余
    if(s==1){    //  if(s%n==1)
     break;
    }
    
   }
   printf("2^%d mod %d = 1\n",i,n);
  }
 }
 return 0;
}