#include<cstdio>
using namespace std;
#define N 16400
#define M 204
#define mod 998244353
int dp[M][N],n,m,fr[N],ifr[N],rev[N],a[N],b[N],ntt[N];
int pw(int a,int p){int as=1;while(p){if(p&1)as=1ll*as*a%mod;a=1ll*a*a%mod;p>>=1;}return as;}
int c(int a,int b){if(a<b)return 0;return 1ll*fr[a]*ifr[b]%mod*ifr[a-b]%mod;}
void dft(int s,int *a,int t)
{
	for(int i=0;i<s;i++)rev[i]=(rev[i>>1]>>1)|(i&1?s>>1:0),ntt[rev[i]]=a[i];
	for(int i=2;i<=s;i<<=1)
	{
		int st=pw(3,(mod-1)/i);
		if(t==-1)st=pw(st,mod-2);
		for(int j=0;j<s;j+=i)
		for(int k=j,t=1;k<j+(i>>1);k++,t=1ll*t*st%mod)
		{
			int v1=ntt[k],v2=1ll*ntt[k+(i>>1)]*t%mod;
			ntt[k]=(v1+v2)%mod;ntt[k+(i>>1)]=(v1-v2+mod)%mod;
		}
	}
	int inv=t==-1?pw(s,mod-2):1;
	for(int i=0;i<s;i++)a[i]=1ll*ntt[i]*inv%mod;
}
int main()
{
	scanf("%d%d",&n,&m);
	dp[0][0]=1;
	fr[0]=ifr[0]=1;for(int i=1;i<=8020;i++)fr[i]=1ll*fr[i-1]*i%mod,ifr[i]=pw(fr[i],mod-2);
	for(int i=1;i<=m;i++)
	{
		int l=1;while(l<=n*2)l<<=1;for(int j=0;j<l;j++)a[j]=b[j]=0;
		for(int j=0;j<=n;j++)a[j]=1ll*dp[i-1][j]*ifr[j]%mod;
		for(int j=1;j<=n;j++)b[j]=ifr[j+2];
		dft(l,a,1);dft(l,b,1);for(int j=0;j<l;j++)a[j]=1ll*a[j]*b[j]%mod;dft(l,a,-1);
		for(int j=0;j<=n;j++)dp[i][j]=(1ll*a[j]*fr[j+2]+1ll*(1ll*j*(j-1)%mod*499122177%mod+j+1)%mod*dp[i-1][j])%mod;
	}
	int as=0;
	for(int j=0;j<=n;j++)as=(as+1ll*dp[m][j]*c(n,j))%mod;
	printf("%d\n",as);
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
using namespace std;
#define N 16400
#define M 204
#define mod 998244353
int dp[M][N],n,m,fr[N],ifr[N],rev[N],a[N],b[N],ntt[N];
int pw(int a,int p){int as=1;while(p){if(p&1)as=1ll*as*a%mod;a=1ll*a*a%mod;p>>=1;}return as;}
int c(int a,int b){if(a<b)return 0;return 1ll*fr[a]*ifr[b]%mod*ifr[a-b]%mod;}
void dft(int s,int *a,int t)
{
	for(int i=0;i<s;i++)rev[i]=(rev[i>>1]>>1)|(i&1?s>>1:0),ntt[rev[i]]=a[i];
	for(int i=2;i<=s;i<<=1)
	{
		int st=pw(3,(mod-1)/i);
		if(t==-1)st=pw(st,mod-2);
		for(int j=0;j<s;j+=i)
		for(int k=j,t=1;k<j+(i>>1);k++,t=1ll*t*st%mod)
		{
			int v1=ntt[k],v2=1ll*ntt[k+(i>>1)]*t%mod;
			ntt[k]=(v1+v2)%mod;ntt[k+(i>>1)]=(v1-v2+mod)%mod;
		}
	}
	int inv=t==-1?pw(s,mod-2):1;
	for(int i=0;i<s;i++)a[i]=1ll*ntt[i]*inv%mod;
}
int main()
{
	scanf("%d%d",&n,&m);
	dp[0][0]=1;
	fr[0]=ifr[0]=1;for(int i=1;i<=8020;i++)fr[i]=1ll*fr[i-1]*i%mod,ifr[i]=pw(fr[i],mod-2);
	for(int i=1;i<=m;i++)
	{
		int l=1;while(l<=n*2)l<<=1;for(int j=0;j<l;j++)a[j]=b[j]=0;
		for(int j=0;j<=n;j++)a[j]=1ll*dp[i-1][j]*ifr[j]%mod;
		for(int j=1;j<=n;j++)b[j]=ifr[j+2];
		dft(l,a,1);dft(l,b,1);for(int j=0;j<l;j++)a[j]=1ll*a[j]*b[j]%mod;dft(l,a,-1);
		for(int j=0;j<=n;j++)dp[i][j]=(1ll*a[j]*fr[j+2]+1ll*(1ll*j*(j-1)%mod*499122177%mod+j+1)%mod*dp[i-1][j])%mod;
	}
	int as=0;
	for(int j=0;j<=n;j++)as=(as+1ll*dp[m][j]*c(n,j))%mod;
	printf("%d\n",as);
}
```