Calculate the number of ways to place n rooks on n × n chessboard so that both following conditions are met:

  * each empty cell is under attack; 
  * exactly k pairs of rooks attack each other. 



An empty cell is under attack if there is at least one rook in the same row or at least one rook in the same column. Two rooks attack each other if they share the same row or column, and there are no other rooks between them. For example, there are only two pairs of rooks that attack each other in the following picture:

<image> One of the ways to place the rooks for n = 3 and k = 2

Two ways to place the rooks are considered different if there exists at least one cell which is empty in one of the ways but contains a rook in another way.

The answer might be large, so print it modulo 998244353.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 200000; 0 ≤ k ≤ (n(n - 1))/(2)).

Output

Print one integer — the number of ways to place the rooks, taken modulo 998244353.

Examples

Input


3 2


Output


6


Input


3 3


Output


0


Input


4 0


Output


24


Input


1337 42


Output


807905441
#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 200005;
const int mod = 998244353;
template <typename _Tp1, typename _Tp2>
__inline__ __attribute__((always_inline)) void add(_Tp1 &a, _Tp2 b) {
  (a += b) >= mod && (a -= mod);
}
template <typename _Tp1, typename _Tp2>
__inline__ __attribute__((always_inline)) void sub(_Tp1 &a, _Tp2 b) {
  (a -= b) < 0 && (a += mod);
}
long long ksm(long long a, long long b = mod - 2) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return res;
}
long long fac[N], inv[N];
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i < N; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  }
  for (int i = 2; i < N; ++i) {
    inv[i] = inv[i - 1] * inv[i] % mod;
  }
}
__inline__ __attribute__((always_inline)) long long C(int n, int m) {
  return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
long long S(int n, int m) {
  long long ans = 0;
  for (int i = 0; i <= m; ++i) {
    if ((m - i) & 1)
      sub(ans, C(m, i) * ksm(i, n) % mod);
    else
      add(ans, C(m, i) * ksm(i, n) % mod);
  }
  ans = ans * inv[m] % mod;
  return ans;
}
int main() {
  init();
  int n;
  long long k;
  read(n, k);
  if (k == 0)
    printf("%lld\n", fac[n]);
  else if (k >= n)
    printf("%lld\n", 0ll);
  else
    printf("%lld\n", 2ll * C(n, n - k) * fac[n - k] % mod * S(n, n - k) % mod);
  return 0;
}
