计数好题

容斥式子：发现只要每个钦定方案的贡献都考虑到再配上容斥系数就是对的

O(n^2)->O(n)

把麻烦的i=0，j=0特殊考虑下

剩下的，先把麻烦的东西化简干净

然后枚举一维i，剩下的二项式定理！！！！

#include
     
      #define reg register int#define il inline#define fi first#define se second#define mk(a,b) make_pair(a,b)#define numb (ch^'0')#define pb push_back#define solid const auto &#define enter cout<
      
       il void rd(T &x){    char ch;x=0;bool fl=false;    while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);    for(x=numb;isdigit(ch=getchar());x=x*10+numb);    (fl==true)&&(x=-x);}template
       
        il void output(T x){
   if(x/10)output(x/10);putchar(x%10+'0');}template
        
         il void ot(T x){
   if(x<0) putchar('-'),x=-x;output(x);putchar(' ');}template
         
          il void prt(T a[],int st,int nd){
   for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');}namespace Miracle{const int N=1e6+5;const int mod=998244353;int ad(int x,int y){    return (x+y)>=mod?x+y-mod:x+y;}int qm(int x,ll y){    int ret=1;    while(y){        if(y&1) ret=(ll)ret*x%mod;x=(ll)x*x%mod;y>>=1;    }return ret;}int inv[N],jie[N];int C(int n,int m){    if(n<0||m<0||n
          
           =0;--i) inv[i]=(ll)inv[i+1]*(i+1)%mod; for(reg i=1;i<=n;++i){ ans=ad(ans,(i+1)&1?mod-(ll)C(n,i)*qm(3,(ll)n*(n-i)+i)%mod:(ll)C(n,i)*qm(3,(ll)n*(n-i)+i)%mod); } ans=ans*2%mod; ll sum=0; ll base=1; for(reg i=0;i<=n-1;++i){ sum=ad(sum,(i+1)&1?mod-(ll)C(n,i)*ad(qm(ad(1,mod-base),n),mod-qm(mod-base,n))%mod:(ll)C(n,i)*ad(qm(ad(1,mod-base),n),mod-qm(mod-base,n))%mod); base=base*3%mod; } ans=(ans+sum*3)%mod; ot(ans); return 0;}}signed main(){ Miracle::main(); return 0;}/* Author: *Miracle**/