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>
using namespace std;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b = b >> 1;
  }
  return res;
}
void extendgcd(long long a, long long b, long long *v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long mminv(long long a, long long b) {
  long long arr[3];
  extendgcd(a, b, arr);
  return arr[0];
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
void swap(long long &x, long long &y) {
  long long temp = x;
  x = y;
  y = temp;
}
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
void google(long long t) { cout << "Case #" << t << ": "; }
vector<long long> sieve(long long n) {
  long long *arr = new long long[n + 1]();
  vector<long long> vect;
  for (long long i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long mod_div(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
long long phin(long long n) {
  long long number = n;
  if (n % 2 == 0) {
    number /= 2;
    while (n % 2 == 0) n /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      number = (number / i * (i - 1));
    }
  }
  if (n > 1) number = (number / n * (n - 1));
  return number;
}
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long N = 200000;
long long fact[N + 1];
long long comb(long long a, long long b) {
  return mod_div(mod_div(fact[a], fact[b], 998244353), fact[a - b], 998244353);
}
int32_t main() {
  c_p_c();
  long long n, k;
  cin >> n >> k;
  if (k >= n) {
    cout << 0 << endl;
    return 0;
  }
  fact[0] = 1;
  for (long long i = 1; i <= n; i++)
    fact[i] = mod_mul(fact[i - 1], i, 998244353);
  long long ans = 0;
  long long place = n - k;
  for (long long i = 0; i <= place; i++) {
    long long sa = comb(place, i);
    sa = mod_mul(sa, expo(place - i, n, 998244353), 998244353);
    if (i & 1)
      ans = mod_sub(ans, sa, 998244353);
    else
      ans = mod_add(ans, sa, 998244353);
  }
  ans = mod_mul(ans, comb(n, place), 998244353);
  if (k != 0) ans = mod_mul(ans, 2, 998244353);
  cout << ans << endl;
}
