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 int inf = 998244353;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto& x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
const int N = 1e6 + 10;
long long fac[N + 1], inv[N + 1], ifac[N + 1];
inline int C(int n, int k) {
  if (k > n) return 0;
  int ans = (fac[n] * 1LL * ifac[k]) % inf;
  return (ans * 1LL * ifac[n - k]) % inf;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  fac[0] = 1;
  ifac[0] = 1;
  inv[1] = 1;
  for (auto i = (1); i <= (N); ++i) fac[i] = (fac[i - 1] * 1LL * i) % inf;
  for (auto i = (2); i <= (N); ++i)
    inv[i] = inf - ((inf / i) * 1LL * inv[inf % i]) % inf;
  for (auto i = (1); i <= (N); ++i)
    ifac[i] = (ifac[i - 1] * 1LL * inv[i]) % inf;
  long long ans = fac[n];
  for (auto i = (1); i <= (n - 1); ++i) {
    long long cur = (fac[n - i] - 1 + inf) % inf;
    cur = (cur * C(n, i)) % inf;
    cur = (cur * fac[i]) % inf;
    cur = (cur + inf) % inf;
    ans = (ans + cur) % inf;
  }
  cout << ans;
  return 0;
}
