As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
for some function 
.
(This equation should hold for any integer x in the range 0 top - 1,
inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
题意：
给出方程f(kx%p)=kf(x)%p ，问在集合A->B上不同的映射函数f 有几种，其中A=B={0,1,2..p-1}，p为素数（除了2），k为小于p的一个常数
解法：
如果K=0，F(0)=0 ，其他的映射关系可以随便选择，方案数为p^(p-1)
对于其他的，将x=k*x带入得，f(k*k*x %p)=k*f(kx%p)%p=k*k*f(x)%p
那么 f(x)%p=k*f(x)%p=k*k*f(x)%p=k*k*k*f(x)%p=......
因为（k^m）%p随着m的增大，会变回k，所以是有一个环在其中的。同时根据费马小定理，m肯定是p-1的因子……因为CF很快且单样例，在代码中直接暴力搞了。
每个环的起始位置有p种可能，共(p-1)/m个环，所以方案数为p^((p-1)/m)
对于K=1特判，因为方程为f(x)=f(x)，每个数都有p种可能，方案数为P^P
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
ll p,k,mod=1e9+7;
ll ksm(ll a,ll x){
ll ans=1;
while(x){
if(x&1)ans=(ans*a)%mod;
x>>=1;
a=(a*a)%mod;
}
return ans;
}
int main(){
scanff(p);
scanff(k);
ll x=1;
ll m=1;
rep(i,1,p){
x=(x*k)%p;
if(x==1)break;
m++;
}
ll ans=0;
if(k==0)ans=ksm(p,p-1);
if(k==1)ans=ksm(p,p);
if(k>=2)ans=ksm(p,(p-1)/m);
printf("%lld\n",ans);
}