Let n be an integer. Consider all permutations on integers 1 to n in lexicographic order, and concatenate them into one big sequence p. For example, if n = 3, then p = [1, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 3, 1, 2, 3, 2, 1]. The length of this sequence will be n ⋅ n!.

Let 1 ≤ i ≤ j ≤ n ⋅ n! be a pair of indices. We call the sequence (p_i, p_{i+1}, ..., p_{j-1}, p_j) a subarray of p. Its length is defined as the number of its elements, i.e., j - i + 1. Its sum is the sum of all its elements, i.e., ∑_{k=i}^j p_k. 

You are given n. Find the number of subarrays of p of length n having sum (n(n+1))/(2). Since this number may be large, output it modulo 998244353 (a prime number). 

Input

The only line contains one integer n (1 ≤ n ≤ 10^6), as described in the problem statement.

Output

Output a single integer — the number of subarrays of length n having sum (n(n+1))/(2), modulo 998244353.

Examples

Input


3


Output


9


Input


4


Output


56


Input


10


Output


30052700

Note

In the first sample, there are 16 subarrays of length 3. In order of appearance, they are:

[1, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 2], [3, 2, 2], [2, 2, 1], [2, 1, 3], [1, 3, 2], [3, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 1], [3, 1, 2], [1, 2, 3], [2, 3, 2], [3, 2, 1]. 

Their sums are 6, 6, 7, 6, 7, 5, 6, 6, 8, 6, 7, 5, 6, 6, 7, 6. As (n(n+1))/(2) = 6, the answer is 9.
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9')
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * w;
}
int n;
int a[1000010], Inv[1000010], fac[1000010];
int main() {
  n = read();
  fac[0] = fac[1] = Inv[0] = Inv[1] = 1;
  for (int i = 2; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % 998244353;
  for (int i = 2; i <= n; i++)
    Inv[i] = 1ll * (998244353 - 998244353 / i) * Inv[998244353 % i] % 998244353;
  for (int i = 2; i <= n; i++) Inv[i] = 1ll * Inv[i - 1] * Inv[i] % 998244353;
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    int tmp = 1ll * (fac[i] - 1 + 998244353) % 998244353;
    int sum = 1ll * tmp * fac[n - i] % 998244353 * fac[n] % 998244353 * Inv[i] %
              998244353 * Inv[n - i] % 998244353;
    ans = 1ll * (ans + sum) % 998244353;
  }
  printf("%d", (ans + 1) % 998244353);
  return 0;
}
