【我也不知道哪里来的题】—尘封的花环（Burnside引理+容斥）

用 k 种颜色对大小为 n 的环进行染色，要求相邻点上的颜色不同，旋转同构的方案视作相同，求本质不同的方案数

先不考虑同构
考虑容斥有几对颜色相同

得到$f(n)=\sum_{i=0}^n(-1)^i{n\choose i}k^{max(n-i,1)}=(k-1)^n+(-1)^n(k-1)$
考虑怎么计算同构

考虑对于旋转$i$位的置换，一个状态会遍历$n/gcd(n,i)$个不同状态
总共就有$gcd(n,i)$个循环
一个不动点就是每个循环的颜色相同的情况
也就是给$f(gcd(n,i))$

$ans=\sum_{i=1}^nf(gcd(n,i))=\sum_{d|n}f(d)\phi(\frac nd)$

由于$1e18$内的数因数只有$1e5$的规模
写个$Pollard-Rho$就可以了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
namespace Get{
	inline ll mul(ll a,ll b,cs ll &mod){
		return (a*b-(ll)((long double)a/mod*b)*mod+mod)%mod;
	}
	inline ll gcd(ll a,ll b){
		return b?gcd(b,a%b):a;
	}
	inline ll ksm(ll a,ll b,cs ll &mod){
		ll res=1;
		for(;b;b>>=1,a=mul(a,a,mod))if(b&1)res=mul(res,a,mod);
		return res;
	}
	inline bool isprime(ll x){
		static int pr[17]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59};
		for(int i=0;i<17;i++)if(x%pr[i]==0)return x==pr[i];
		if(x<pr[16])return false;
		ll t=x-1,s=0;
		while(!(t&1))t>>=1,s++;
		for(int i=0;i<17;i++){
			ll a=pr[i],b=ksm(a,t,x);
			for(int j=1;j<=s;j++){
				ll k=mul(b,b,x);
				if(k==1&&b!=1&&b!=x-1)return false;
				b=k;
			}
			if(b!=1)return false;
		}
		return true;
	}
	inline ll f(ll x,ll c,cs ll &mod){
		return (mul(x,x,mod)+c)%mod;
	}
	inline ll pro(ll x){
		ll s=0,t=0,c=rand()%(x-1)+1;
		for(int goal=1;;goal<<=1,s=t){
			ll val=1;
			for(int step=1;step<=goal;step++){
				t=f(t,c,x);
				val=mul(val,abs(t-s),x);
				if(step%127==0){
					ll d=gcd(val,x);
					if(d>1)return d;
				}
			}
			ll d=gcd(val,x);
			if(d>1)return d;
		}
	}
}
cs int mod=998244353;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
ll pr[100],n;
int cnt[100],tot,k;
inline void getfac(ll x){
	if(x==1)return;
	if(Get::isprime(x)){pr[++tot]=x;return;}
	ll p=x;
	while(p>=x)p=Get::pro(x);
	while(x%p==0)x/=p;
	getfac(x),getfac(p);
}
int ans;
inline void calc(ll d,ll phi){
	d=n/d;
	int res=ksm(k,d%(mod-1));
	(d&1)?Dec(res,k):Add(res,k);
	Add(ans,mul(phi%mod,res));
}
void dfs(int pos,ll prod,ll phi){
	if(pos>tot)return calc(prod,phi);
	dfs(pos+1,prod,phi);
	for(int i=1;i<=cnt[pos];i++){
		prod*=pr[pos],phi*=(pr[pos]-(i==1));
		dfs(pos+1,prod,phi);
	}
}
signed main(){
	int T=read();
	while(T--){
		scanf("%lld",&n),k=(read()-1)%mod;
		tot=ans=0;
		getfac(n);
		memset(cnt,0,sizeof(cnt));
		sort(pr+1,pr+tot+1);
		tot=unique(pr+1,pr+tot+1)-pr-1;
		for(int i=1;i<=tot;i++){
			ll x=n;
			while(x%pr[i]==0)x/=pr[i],cnt[i]++;
		}
		dfs(1,1,1);
		cout<<mul(ans,Inv(n%mod))<<'\n';
	}
}