#include<cstdio>
typedef long long ll;
const int N=8024,D=203,M=998244353,G=3,K=14;
inline int Pow(int a,int m){int s=1;for(;m;m>>=1)m&1?s=(ll)s*a%M:0,a=(ll)a*a%M;return s;}
int n,m,f[N],fac[N],invf[N],ans,a[1<<K],b[1<<K],rev[1<<K];
inline int C(int n,int m){
	if(n<m||m<0)return 0;
	return(ll)fac[n]*invf[m]%M*invf[n-m]%M;
}
inline void Init(){
	for(int i=0;i<(1<<K);i++)rev[i]=rev[i>>1]>>1|(i&1)<<K-1;
}
inline void Ntt(int*a,int f){
	int tmp,s,w;
	for(int i=0;i<(1<<K);i++)if(i<rev[i])
	  tmp=a[i],a[i]=a[rev[i]],a[rev[i]]=tmp;
	for(int k=1;k<(1<<K);k<<=1){
	  w=Pow(G,f>0?M/2/k:M-1-M/2/k);
	  for(int i=0;i<(1<<K);i+=k+k){
		s=1;
		for(int j=0;j<k;j++,s=(ll)s*w%M){
		  tmp=(ll)s*a[i+k+j]%M;
		  (a[i+k+j]=a[i+j]-tmp)<0&&(a[i+k+j]+=M);
		  (a[i+j]+=tmp)>=M&&(a[i+j]-=M);
		}
	  }
	}
	if(f<0){
	  tmp=Pow(1<<K,M-2);
	  for(int i=0;i<(1<<K);i++)a[i]=(ll)a[i]*tmp%M;
	}
}
int main(){
	fac[0]=1;
	for(int i=1;i<N;i++)fac[i]=(ll)fac[i-1]*i%M;
	invf[N-1]=Pow(fac[N-1],M-2);
	for(int i=N-1;i;i--)invf[i-1]=(ll)invf[i]*i%M;
	scanf("%d%d",&n,&m);
	Init();
	for(int i=1;i<=n;i++)b[i]=invf[i+2];
	Ntt(b,1);
	f[0]=1;
	for(int j=1;j<=m;j++){
	  for(int i=0;i<(1<<K);i++)a[i]=0;
	  for(int i=0;i<=n;i++)a[i]=(ll)f[i]*invf[i]%M;
	  Ntt(a,1);
	  for(int i=0;i<(1<<K);i++)a[i]=(ll)a[i]*b[i]%M;
	  Ntt(a,-1);
	  for(int i=0;i<=n;i++)f[i]=((ll)f[i]*(C(i,2)+i+1)+(ll)fac[i+2]*a[i])%M;
	}
	for(int k=0;k<=n;k++)
	  ans=(ans+(ll)f[k]*C(n,k))%M;
	printf("%d\n",ans);
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

We have an N \times M grid. The square at the i-th row and j-th column will be denoted as (i,j). Particularly, the top-left square will be denoted as (1,1), and the bottom-right square will be denoted as (N,M). Takahashi painted some of the squares (possibly zero) black, and painted the other squares white.

We will define an integer sequence A of length N, and two integer sequences B and C of length M each, as follows:

* A_i(1\leq i\leq N) is the minimum j such that (i,j) is painted black, or M+1 if it does not exist.
* B_i(1\leq i\leq M) is the minimum k such that (k,i) is painted black, or N+1 if it does not exist.
* C_i(1\leq i\leq M) is the maximum k such that (k,i) is painted black, or 0 if it does not exist.



How many triples (A,B,C) can occur? Find the count modulo 998244353.

Constraints

* 1 \leq N \leq 8000
* 1 \leq M \leq 200
* N and M are integers.



Partial Score

* 1500 points will be awarded for passing the test set satisfying N\leq 300.

Constraints

* 1 \leq N \leq 8000
* 1 \leq M \leq 200
* N and M are integers.

Input

Input is given from Standard Input in the following format:


N M

Output

Print the number of triples (A,B,C), modulo 998244353.

Examples

Input

2 3


Output

64


Input

4 3


Output

2588


Input

17 13


Output

229876268


Input

5000 100


Output

57613837

### Response

```cpp
#include<cstdio>
typedef long long ll;
const int N=8024,D=203,M=998244353,G=3,K=14;
inline int Pow(int a,int m){int s=1;for(;m;m>>=1)m&1?s=(ll)s*a%M:0,a=(ll)a*a%M;return s;}
int n,m,f[N],fac[N],invf[N],ans,a[1<<K],b[1<<K],rev[1<<K];
inline int C(int n,int m){
	if(n<m||m<0)return 0;
	return(ll)fac[n]*invf[m]%M*invf[n-m]%M;
}
inline void Init(){
	for(int i=0;i<(1<<K);i++)rev[i]=rev[i>>1]>>1|(i&1)<<K-1;
}
inline void Ntt(int*a,int f){
	int tmp,s,w;
	for(int i=0;i<(1<<K);i++)if(i<rev[i])
	  tmp=a[i],a[i]=a[rev[i]],a[rev[i]]=tmp;
	for(int k=1;k<(1<<K);k<<=1){
	  w=Pow(G,f>0?M/2/k:M-1-M/2/k);
	  for(int i=0;i<(1<<K);i+=k+k){
		s=1;
		for(int j=0;j<k;j++,s=(ll)s*w%M){
		  tmp=(ll)s*a[i+k+j]%M;
		  (a[i+k+j]=a[i+j]-tmp)<0&&(a[i+k+j]+=M);
		  (a[i+j]+=tmp)>=M&&(a[i+j]-=M);
		}
	  }
	}
	if(f<0){
	  tmp=Pow(1<<K,M-2);
	  for(int i=0;i<(1<<K);i++)a[i]=(ll)a[i]*tmp%M;
	}
}
int main(){
	fac[0]=1;
	for(int i=1;i<N;i++)fac[i]=(ll)fac[i-1]*i%M;
	invf[N-1]=Pow(fac[N-1],M-2);
	for(int i=N-1;i;i--)invf[i-1]=(ll)invf[i]*i%M;
	scanf("%d%d",&n,&m);
	Init();
	for(int i=1;i<=n;i++)b[i]=invf[i+2];
	Ntt(b,1);
	f[0]=1;
	for(int j=1;j<=m;j++){
	  for(int i=0;i<(1<<K);i++)a[i]=0;
	  for(int i=0;i<=n;i++)a[i]=(ll)f[i]*invf[i]%M;
	  Ntt(a,1);
	  for(int i=0;i<(1<<K);i++)a[i]=(ll)a[i]*b[i]%M;
	  Ntt(a,-1);
	  for(int i=0;i<=n;i++)f[i]=((ll)f[i]*(C(i,2)+i+1)+(ll)fac[i+2]*a[i])%M;
	}
	for(int k=0;k<=n;k++)
	  ans=(ans+(ll)f[k]*C(n,k))%M;
	printf("%d\n",ans);
	return 0;
}
```