To AmShZ, all arrays are equal, but some arrays are more-equal than others. Specifically, the arrays consisting of n elements from 1 to n that can be turned into permutations of numbers from 1 to n by adding a non-negative integer to each element.

Mashtali who wants to appear in every problem statement thinks that an array b consisting of k elements is compatible with a more-equal array a consisting of n elements if for each 1 ≤ i ≤ k we have 1 ≤ b_i ≤ n and also a_{b_1} = a_{b_2} = … = a_{b_k}.

Find the number of pairs of arrays a and b such that a is a more-equal array consisting of n elements and b is an array compatible with a consisting of k elements modulo 998244353.

Note that the elements of b are not necessarily distinct, same holds for a.

Input

The first line of input contains two integers n and k (1 ≤ n ≤ 10^9 , 1 ≤ k ≤ 10^5).

Output

Print a single integer — the answer to the problem modulo 998244353.

Examples

Input


1 1


Output


1


Input


2 2


Output


8


Input


5 4


Output


50400


Input


20 100


Output


807645526


Input


10000000 10000


Output


883232350

Note

There are eight possible pairs for the second example: 

  1. a = \{1, 1\}, b = \{1, 1\} 
  2. a = \{1, 1\}, b = \{1, 2\} 
  3. a = \{1, 1\}, b = \{2, 1\} 
  4. a = \{1, 1\}, b = \{2, 2\} 
  5. a = \{1, 2\}, b = \{1, 1\} 
  6. a = \{1, 2\}, b = \{2, 2\} 
  7. a = \{2, 1\}, b = \{1, 1\} 
  8. a = \{2, 1\}, b = \{2, 2\} 
#include <cstdio>
#include <iostream>
using namespace std;
const int M = 300005;
const int MOD = 998244353;
#define int long long
int read()
{
	int x=0,f=1;char c;
	while((c=getchar())<'0' || c>'9') {if(c=='-') f=-1;}
	while(c>='0' && c<='9') {x=(x<<3)+(x<<1)+(c^48);c=getchar();}
	return x*f;
}
namespace poly
{
	int len,A[M],B[M],rev[M];
	int qkpow(int a,int b)
	{
		int r=1;
		while(b>0)
		{
			if(b&1) r=r*a%MOD;
			a=a*a%MOD;
			b>>=1;
		}
		return r;
	}
	void NTT(int *a,int len,int op)
	{
		for(int i=0;i<len;i++)
		{
			rev[i]=(rev[i>>1]>>1)|((len/2)*(i&1));
			if(i<rev[i]) swap(a[i],a[rev[i]]);
		}
		for(int s=2;s<=len;s<<=1)
		{
			int t=s/2,w=(op==1)?qkpow(3,(MOD-1)/s):qkpow(3,MOD-1-(MOD-1)/s);
			for(int i=0;i<len;i+=s)
				for(int j=0,x=1;j<t;j++,x=x*w%MOD)
				{
					int fe=a[i+j],fo=a[i+j+t];
					a[i+j]=(fe+x*fo)%MOD;
					a[i+j+t]=((fe-x*fo)%MOD+MOD)%MOD;
				}
		}
		if(op==1) return ;
		int inv=qkpow(len,MOD-2);
		for(int i=0;i<len;i++) a[i]=a[i]*inv%MOD;
	}
	void work(int n,int m,int *a,int *b,int *c)
	{
		len=1;while(len<n+m) len<<=1;
		for(int i=0;i<len;i++) A[i]=B[i]=0;
		for(int i=0;i<n;i++) A[i]=a[i];
		for(int i=0;i<m;i++) B[i]=b[i];
		NTT(A,len,1);NTT(B,len,1);
		for(int i=0;i<len;i++) A[i]=A[i]*B[i]%MOD;
		NTT(A,len,-1);
		for(int i=0;i<n+m;i++) c[i]=A[i];
	}
}
int n,k,ans,fac[M],inv[M],a[M],b[M],c[M];
void init(int n)
{
	inv[0]=inv[1]=fac[0]=1;
	for(int i=2;i<=n;i++) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=2;i<=n;i++) inv[i]=inv[i]*inv[i-1]%MOD;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%MOD;
}
signed main()
{
	init(1e5);
	n=read();k=read();c[0]=1;
	for(int i=0;i<=k;i++)
	{
		a[i]=poly::qkpow(i,k)*inv[i]%MOD;
		b[i]=(i%2)?(MOD-inv[i]):inv[i];
		if(i>0) c[i]=c[i-1]*(n-i+1)%MOD*poly::qkpow(i,MOD-2)%MOD;
	}
	poly::work(k+1,k+1,a,b,a);
	for(int i=0;i<=k;i++)
		ans=(ans+a[i]*c[i]%MOD*poly::qkpow(n+1,n-i)%MOD*fac[i])%MOD;
	printf("%lld\n",ans);
}