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;

static struct FastInput {
  static constexpr int BUF_SIZE = 1 << 20;
  char buf[BUF_SIZE];
  size_t chars_read = 0;
  size_t buf_pos = 0;
  FILE *in = stdin;
  char cur = 0;
 
  inline char get_char() {
    if (buf_pos >= chars_read) {
      chars_read = fread(buf, 1, BUF_SIZE, in);
      buf_pos = 0;
      buf[0] = (chars_read == 0 ? -1 : buf[0]);
    }
    return cur = buf[buf_pos++];
  }
 
  inline void tie(int) {}
 
  inline explicit operator bool() {
    return cur != -1;
  }
 
  inline static bool is_blank(char c) {
    return c <= ' ';
  }
 
  inline bool skip_blanks() {
    while (is_blank(cur) && cur != -1) {
      get_char();
    }
    return cur != -1;
  }
 
  inline FastInput& operator>>(char& c) {
    skip_blanks();
    c = cur;
    return *this;
  }
  
  inline FastInput& operator>>(string& s) {
    if (skip_blanks()) {
      s.clear();
      do {
        s += cur;
      } while (!is_blank(get_char()));
    }
    return *this;
  }
 
  template <typename T>
  inline FastInput& read_integer(T& n) {
    // unsafe, doesn't check that characters are actually digits
    n = 0;
    if (skip_blanks()) {
      int sign = +1;
      if (cur == '-') {
        sign = -1;
        get_char();
      }
      do {
        n += n + (n << 3) + cur - '0';
      } while (!is_blank(get_char()));
      n *= sign;
    }
    return *this;
  }
 
  template <typename T>
  inline typename enable_if<is_integral<T>::value, FastInput&>::type operator>>(T& n) {
    return read_integer(n);
  }
  
  #if !defined(_WIN32) || defined(_WIN64)
  inline FastInput& operator>>(__int128& n) {
    return read_integer(n);
  }
  #endif
 
  template <typename T>
  inline typename enable_if<is_floating_point<T>::value, FastInput&>::type operator>>(T& n) {
    // not sure if really fast, for compatibility only
    n = 0;
    if (skip_blanks()) {
      string s;
      (*this) >> s;
      sscanf(s.c_str(), "%lf", &n);
    }
    return *this;
  }
} fast_input;
 
#define cin fast_input

static struct FastOutput {
  static constexpr int BUF_SIZE = 1 << 20;
  char buf[BUF_SIZE];
  size_t buf_pos = 0;
  static constexpr int TMP_SIZE = 1 << 20;
  char tmp[TMP_SIZE];
  FILE *out = stdout;
 
  inline void put_char(char c) {
    buf[buf_pos++] = c;
    if (buf_pos == BUF_SIZE) {
      fwrite(buf, 1, buf_pos, out);
      buf_pos = 0;
    }
  }
 
  ~FastOutput() {
    fwrite(buf, 1, buf_pos, out);
  }
 
  inline FastOutput& operator<<(char c) {
    put_char(c);
    return *this;
  }
 
  inline FastOutput& operator<<(const char* s) {
    while (*s) {
      put_char(*s++);
    }
    return *this;
  }
 
  inline FastOutput& operator<<(const string& s) {
    for (int i = 0; i < (int) s.size(); i++) {
      put_char(s[i]);
    }
    return *this;
  }
 
  template <typename T>
  inline char* integer_to_string(T n) {
    // beware of TMP_SIZE
    char* p = tmp + TMP_SIZE - 1;
    if (n == 0) {
      *--p = '0';
    } else {
      bool is_negative = false;
      if (n < 0) {
        is_negative = true;
        n = -n;
      }
      while (n > 0) {
        *--p = (char) ('0' + n % 10);
        n /= 10;
      }
      if (is_negative) {
        *--p = '-';
      }
    }
    return p;
  }
 
  template <typename T>
  inline typename enable_if<is_integral<T>::value, char*>::type stringify(T n) {
    return integer_to_string(n);
  }
 
  #if !defined(_WIN32) || defined(_WIN64)
  inline char* stringify(__int128 n) {
    return integer_to_string(n);
  }
  #endif
 
  template <typename T>
  inline typename enable_if<is_floating_point<T>::value, char*>::type stringify(T n) {
    sprintf(tmp, "%.17f", n);
    return tmp;
  }
 
  template <typename T>
  inline FastOutput& operator<<(const T& n) {
    auto p = stringify(n);
    for (; *p != 0; p++) {
      put_char(*p);
    }
    return *this;
  }
} fast_output;

// here puts define
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

#define rint register int
#define rll register ll 
#define pii pair<int, int>
#define pll pair<ll, ll>
#define fors(i, a, b) for (ll i = (a); i <= (b); ++i)
#define _fors(i, a, b) for (ll i = (a); i >= (b); --i)
#define cyc(m) fors(rqtwqtqwt, 1, m)
#define mp(a, b) make_pair(a, b)
#define mt(a, b, c) make_tuple(a, b, c)
#define mem(A, b) memset(A, b, sizeof(A))
#define all(X) (X).begin(), (X).end()
#define y0 gawgfawgawg
#define y1 sdfyseyegeh
#define pb push_back
#define eb emplace_back
#define cout fast_output
#define endl '\n'
#define yes cout << "YES" << endl
#define no cout << "NO" << endl

int start_time;

const int mod = 998244353;

inline int add(int a, int b) {
  return (((a + b) % mod) + mod) % mod;
}

inline int mul(int a, int b) {
  return (((1LL * a * b) % mod) + mod) % mod;
}

inline int qpow(int a, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = mul(ret, a);
    a = mul(a, a);
    p >>= 1;
  }
  return ret;
}

const int _ = 1e5 + 5;

int inv[_], itab[_];

int cda[4*_], cdb[4*_];
int res[4*_];

inline void NTT(int* a, int a_size, bool inverse) {
  int n = a_size;
  // 原地快速bit reversal
  for(int i = 0, j = 0; i < n; i++) {
    if(j > i) swap(a[i], a[j]);
    int k = n;
    while(j & (k >>= 1)) j &= ~k;
    j |= k;
  }

  int root = inverse ? inv[3] : 3;

  for(int step = 1; step < n; step <<= 1) {
    int alpha = (mod-1) / step / 2;
    // 为求高效，我们并不是依次执行各个完整的DFT合并，而是枚举下标k
    // 对于一个下标k，执行所有DFT合并中该下标对应的蝴蝶操作，即通过E[k]和O[k]计算X[k]
    int omega = qpow(root, alpha), omegak = 1;
    for(int k = 0; k < step; k++) {
      // 计算omega^k. 这个方法效率低，但如果用每次乘omega的方法递推会有精度问题。
      // 有更快更精确的递推方法，为了清晰起见这里略去
      for(int Ek = k; Ek < n; Ek += step << 1) { // Ek是某次DFT合并中E[k]在原始序列中的下标
        int Ok = Ek + step; // Ok是该DFT合并中O[k]在原始序列中的下标
        int t = mul(a[Ok], omegak); // 蝴蝶操作：x1 * omega^k
        a[Ok] = add(a[Ek], -t);  // 蝴蝶操作：y1 = x0 - t
        a[Ek] = add(a[Ek], t);         // 蝴蝶操作：y0 = x0 + t
      }
      omegak = mul(omegak, omega);
    }
  }

  if(inverse) {
    int invn = qpow(n, mod - 2);
    fors(i, 0, n-1) a[i] = mul(a[i], invn);
  }

  return ;
}

// 用FFT实现的快速多项式乘法
inline int NTT_mul(int* v1, int v1_size, int* v2, int v2_size) {
  int s1 = v1_size, s2 = v2_size, S = 2;
  while(S < s1 + s2) S <<= 1;
  fors(i, 0, S-1) cda[i] = 0, cdb[i] = 0;
  for(int i = 0; i < s1; i++) cda[i] = v1[i];
  NTT(cda, S, false);
  for(int i = 0; i < s2; i++) cdb[i] = v2[i];
  NTT(cdb, S, false);
  for(int i = 0; i < S; i++) cda[i] = mul(cda[i], cdb[i]);
  NTT(cda, S, true);
  for(int i = 0; i < s1 + s2 - 1; i++) res[i] = cda[i]; // 虚部均为0
  return s1 + s2 - 1;
}

int n, k;
int a[_], b[_];

void solve() {
  cin >> n >> k;
  inv[1] = 1;
  fors(i, 2, max(3, k)) inv[i] = mul(-mod/i, inv[mod%i]);
  itab[0] = 1;
  fors(i, 1, k) itab[i] = mul(itab[i-1], inv[i]);
  fors(i, 0, k) a[i] = i % 2 ? mod-itab[i] : itab[i];
  fors(i, 1, k) b[i] = mul(qpow(i, k), itab[i]);
  NTT_mul(a, k + 1, b, k + 1);
  int c = 1, iv = qpow((n + 1) % mod, mod - 2), d = 1;
  int ans = 0;
  fors(i, 0, k) {
    ans = add(ans, mul(mul(c, d), res[i]));
    c = mul(c, n - i);
    d = mul(d, iv);
  }
  ans = mul(ans, qpow(n + 1, n));
  cout << ans << endl;
  return ;
}

signed main() {
  #ifdef Sakuyalove
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  #endif 
  ios::sync_with_stdio(false);
  cin.tie(0);
  start_time = clock();
  int T = 1;
  // cin >> T;
  while (T--) {
    solve();
  }

  #ifdef Sakuyalove
    cout << "time = " << clock() - start_time << endl;
  #endif

  return 0;
}