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>
#pragma GCC optimize("O3")
using namespace std;
long long int mod = 998244353LL;
long long int factmod[1000100];
unsigned long long powmod(long long z, long long a) {
  if (a == 0) return 1;
  if (a == 1) return z;
  long long x = powmod(z, a / 2);
  x = (x * x) % mod;
  if (a % 2 == 1) x = (x * z) % mod;
  return x;
}
unsigned long long invmod(long long z) { return powmod(z, mod - 2); }
int main() {
  long long int n;
  scanf("%lld", &n);
  factmod[0] = 1;
  factmod[1] = 1;
  for (int x = 2; x <= n + 10; x += 1) {
    factmod[x] = factmod[x - 1] * x % mod;
  }
  long long int b = 0;
  for (long long k = 1; k < n; k++) {
    long long int a = (k * k) % mod;
    a *= (factmod[n] * invmod(factmod[k + 1]) % mod);
    a %= mod;
    a *= invmod(factmod[n - k - 1]);
    a %= mod;
    a *= factmod[n - k - 1];
    b += a;
    b %= mod;
  }
  long long int c = factmod[n] * n - n + 1;
  long long int ans = c - b + mod;
  ans %= mod;
  printf("%lld\n", ans);
  return 0;
}
