这道题

求助各位大佬，tle了，时间复杂度也正确，已使出全身解数都没有卡过，求助各位大佬。

代码：

#pragma GCC optimize(3)
#pragma GCC target("avx")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define int __int128
using namespace std;
const int p=998244353;
int n,k,f[260][260];
int c[260][260];
int sum[260][260];
inline int read() {
	int X=0;
	bool flag=1;
	char ch=getchar();
	while(ch<'0'||ch>'9') {
		if(ch=='-') flag=0;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9') {
		X=(X<<1)+(X<<3)+ch-'0';
		ch=getchar();
	}
	if(flag) return X;
	return ~(X-1);
}
inline void write(int X) {
	if(X<0) {
		X=~(X-1);
		putchar('-');
	}
	if(X>9) write(X/10);
	putchar(X%10+'0');
}

inline int qpow(int x,int y) {
	int r=1;
	while(y) {
		if(y&1) r=r*x%p;
		x=x*x%p,y>>=1;
	}
	return r;
}
signed main() {
	n=read();
	k=read();
	f[1][0]=1;
	for(register int i=0; i<=k; ++i) {
		sum[1][i]=1;
	}
	for(register int i=0; i<=n; ++i) {
		c[i][0]=1;
	}
	for(register int i=1; i<=n; ++i) {
		for(register int j=1; j<=i; ++j) {
			c[i][j]=(c[i-1][j-1]+c[i-1][j])%p;
		}
	}
	for(register int i=2; i<=n; ++i) {
		for(register int j=1; j<=k; ++j) {
			for(register int t=1; t<=i-1; t+=10) {
				f[i][j]=(f[i][j]+c[n-t][i-t]*sum[t][j-1]%p*qpow(k-j+1,(i-2+t-1)*(i-t)/2)%p)%p;//原始式子是这个，下面是循环展开 
				if(t+1<=i-1) f[i][j]=(f[i][j]+c[n-(t+1)][i-(t+1)]*sum[(t+1)][j-1]%p*qpow(k-j+1,(i-2+(t+1)-1)*(i-(t+1))/2)%p)%p;
				if(t+2<=i-1) f[i][j]=(f[i][j]+c[n-(t+2)][i-(t+2)]*sum[(t+2)][j-1]%p*qpow(k-j+1,(i-2+(t+2)-1)*(i-(t+2))/2)%p)%p;
				if(t+3<=i-1) f[i][j]=(f[i][j]+c[n-(t+3)][i-(t+3)]*sum[(t+3)][j-1]%p*qpow(k-j+1,(i-2+(t+3)-1)*(i-(t+3))/2)%p)%p;
				if(t+4<=i-1) f[i][j]=(f[i][j]+c[n-(t+4)][i-(t+4)]*sum[(t+4)][j-1]%p*qpow(k-j+1,(i-2+(t+4)-1)*(i-(t+4))/2)%p)%p;
				if(t+5<=i-1) f[i][j]=(f[i][j]+c[n-(t+5)][i-(t+5)]*sum[(t+5)][j-1]%p*qpow(k-j+1,(i-2+(t+5)-1)*(i-(t+5))/2)%p)%p;
				if(t+6<=i-1) f[i][j]=(f[i][j]+c[n-(t+6)][i-(t+6)]*sum[(t+6)][j-1]%p*qpow(k-j+1,(i-2+(t+6)-1)*(i-(t+6))/2)%p)%p;
				if(t+7<=i-1) f[i][j]=(f[i][j]+c[n-(t+7)][i-(t+7)]*sum[(t+7)][j-1]%p*qpow(k-j+1,(i-2+(t+7)-1)*(i-(t+7))/2)%p)%p;
				if(t+8<=i-1) f[i][j]=(f[i][j]+c[n-(t+8)][i-(t+8)]*sum[(t+8)][j-1]%p*qpow(k-j+1,(i-2+(t+8)-1)*(i-(t+8))/2)%p)%p;
				if(t+9<=i-1) f[i][j]=(f[i][j]+c[n-(t+9)][i-(t+9)]*sum[(t+9)][j-1]%p*qpow(k-j+1,(i-2+(t+9)-1)*(i-(t+9))/2)%p)%p;
			}
			sum[i][j]=(sum[i][j-1]+f[i][j])%p;
		}
	}
	int ans=0;
	for(register int i=1; i<=k; ++i) {
		ans=(ans+f[n][i])%p;
	}
	write(ans);
	return 0;
}