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 <bits/stdc++.h>
 
using namespace std;
 
typedef long long                                        ll;
typedef long double                                      ld;
typedef pair <int, int>                                  pii;
typedef pair <pii, int>                                  ppi;
typedef pair <int, pii>                                  pip;
typedef pair <pii, pii>                                  ppp;
typedef pair <ll, ll>                                    pll;
 
# define A                                               first
# define B                                               second
# define endl                                            '\n'
# define sep                                             ' '
# define all(x)                                          x.begin(), x.end()
# define kill(x)                                         return cout << x << endl, 0
# define SZ(x)                                           int(x.size())
# define lc                                              id << 1
# define rc                                              id << 1 | 1
# define InTheNameOfGod                                  ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
 
ll power(ll a, ll b, ll md) {return (!b ? 1 : (b & 1 ? a * power(a * a % md, b / 2, md) % md : power(a * a % md, b / 2, md) % md));}
 
const int xn = 3e5 + 10;
const int xm = 18;
const int sq = 320;
const int inf = 1e9 + 10;
const ll INF = 1e18 + 10;
const int mod = 998244353;
const int base = 257;
 
int n, k, C[xn], ans, A[xn], B[xn], rev[xn];
int fact[xn], ifact[xn];
 
void NTT(int *A, bool inv){
	int n = (1 << xm);
	for (int i = 0; i < (1 << xm); ++ i)
		if (rev[i] < i)
			swap(A[i], A[rev[i]]);
	for (int ln = 1; ln < n; ln <<= 1){
		int w = power(3, mod / 2 / ln, mod);
		if (inv)
			w = power(w, mod - 2, mod);
		for (int i = 0; i < n; i += ln + ln){
			int wn = 1;
			for (int j = i; j < i + ln; ++ j){
				int x = A[j], y = 1ll * A[j + ln] * wn % mod;
				A[j] = (x + y) % mod;
				A[j + ln] = (x - y + mod) % mod;
				wn = 1ll * wn * w % mod;
			}
		}
	}
	if (inv){
		int invn = power(1 << xm, mod - 2, mod);
		for (int i = 0; i < n; ++ i)
			A[i] = 1ll * A[i] * invn % mod;
	}
}
 
int main(){
	InTheNameOfGod;
	
	cin >> n >> k;
	C[0] = fact[0] = 1;
	for (int i = 1; i < xn; ++ i){
		C[i] = 1ll * C[i - 1] * (n - i + 1) % mod * power(i, mod - 2, mod) % mod;
		fact[i] = 1ll * fact[i - 1] * i % mod;
	}
	ifact[xn - 1] = power(fact[xn - 1], mod - 2, mod);
	for (int i = xn - 2; i >= 0; -- i)
		ifact[i] = 1ll * ifact[i + 1] * (i + 1) % mod;
	for (int i = 1; i < (1 << xm); ++ i)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (xm - 1));
	for (int i = 0; i <= k; ++ i){
		A[i] = 1ll * power(i, k, mod) * ifact[i] % mod;
		B[i] = ifact[i];
		if (i % 2)
			B[i] = (mod - B[i]) % mod;
	}
	NTT(A, 0), NTT(B, 0);
	for (int i = 0; i < xn; ++ i)
		A[i] = 1ll * A[i] * B[i] % mod;
	NTT(A, 1);
	for (int i = 1; i <= k; ++ i)
		ans = (ans + 1LL * A[i] * C[i] % mod * power(n + 1, n - i, mod) % mod * fact[i] % mod) % mod;
	cout << ans << endl;
 
	return 0;
}