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;
const long long N = 1e6 + 5;
const long long MOD = 998244353;
long long n, ans = 0;
long long fact[N], invfact[N];
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long modinv(long long k) { return pow(k, MOD - 2, MOD); }
void precompute() {
  fact[0] = fact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= MOD;
  }
  invfact[N - 1] = modinv(fact[N - 1]);
  for (long long i = N - 2; i >= 0; i--) {
    invfact[i] = invfact[i + 1] * (i + 1);
    invfact[i] %= MOD;
  }
}
long long nCr(long long x, long long y) {
  if (y > x) return 0;
  long long num = fact[x];
  num *= invfact[y];
  num %= MOD;
  num *= invfact[x - y];
  num %= MOD;
  return num;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  precompute();
  cin >> n;
  long long ans = fact[n] * n;
  ans %= MOD;
  for (long long k = 1; k <= n; k++) {
    long long cur = k * k;
    cur %= MOD;
    cur *= nCr(n, k + 1);
    cur %= MOD;
    cur *= fact[n - k - 1];
    cur %= MOD;
    ans -= cur;
    ans += MOD;
    ans %= MOD;
  }
  ans -= (n - 1);
  ans += MOD;
  ans %= MOD;
  cout << ans;
  return 0;
}
