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

當n=1||n為偶數時，一定不存在解...

當n為奇數時，一定存在解....

然後用最普通的累成取模就可以過.....

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int n;
int gcd (int a,int b)
{
    return b==0? a:gcd(b,a%b);
}
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        if(n==1||n%2==0)
            printf("2^? mod %d = 1\n",n);
        else
        {
            int ans=1,m=2;
            while (m%n!=1)
            {
                m=(m*2)%n;
                ans++;
            }
            printf("2^%d mod %d = 1\n",ans,n);
        }
    }
    return 0;
}