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() {
  char c = getchar();
  int p = 1, ret = 0;
  while ((c < '0') || (c > '9')) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while ((c >= '0') && (c <= '9'))
    ret = (ret << 1) + (ret << 3) + c - '0', c = getchar();
  return ret * p;
}
int a[1000005];
int jc[1000005], ny[1000005];
int n, ans;
inline int pow(int x, int k) {
  int ans = 1;
  for (; k; k >>= 1, x = 1LL * x * x % 998244353)
    if (k & 1) ans = 1LL * ans * x % 998244353;
  return ans;
}
int main() {
  n = read();
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = 1LL * jc[i - 1] * i % 998244353;
  ny[n] = pow(jc[n], 998244353 - 2);
  for (int i = n; i; i--) ny[i - 1] = 1LL * ny[i] * i % 998244353;
  for (int i = 1; i <= n; i++) a[i] = 1LL * jc[n] * ny[i] % 998244353;
  for (int i = 1; i <= n; i++) a[i] = (a[i] - a[i + 1] + 998244353) % 998244353;
  for (int i = 1; i < n; i++) ans = (ans - 1LL * a[i] * i) % 998244353;
  ans = (ans + 1LL * jc[n] * n) % 998244353;
  ans = (ans - n + 1 + 998244353) % 998244353;
  if (ans < 0) ans += 998244353;
  printf("%d", ans);
}
