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;
template <class T, size_t N>
ostream& operator<<(ostream& os, const array<T, N>& v);
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  bool f = 1;
  os << "[";
  for (auto& i : v) {
    if (!f) os << ", ";
    os << i;
    f = 0;
  }
  return os << "]";
}
template <class T, size_t N>
ostream& operator<<(ostream& os, const array<T, N>& v) {
  bool f = 1;
  os << "[";
  for (auto& i : v) {
    if (!f) os << ", ";
    os << i;
    f = 0;
  }
  return os << "]";
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const set<T1, T2>& v) {
  bool f = 1;
  os << "[";
  for (auto& i : v) {
    if (!f) os << ", ";
    os << i;
    f = 0;
  }
  return os << "]";
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const multiset<T1, T2>& v) {
  bool f = 1;
  os << "[";
  for (auto& i : v) {
    if (!f) os << ", ";
    os << i;
    f = 0;
  }
  return os << "]";
}
template <class T1, class T2, class T3>
ostream& operator<<(ostream& os, const map<T1, T2, T3>& v) {
  bool f = 1;
  os << "[";
  for (auto& ii : v) {
    if (!f) os << ", ";
    os << "(" << ii.first << " -> " << ii.second << ") ";
    f = 0;
  }
  return os << "]";
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const multimap<T1, T2>& v) {
  bool f = 1;
  os << "[";
  for (auto& ii : v) {
    if (!f) os << ", ";
    os << "(" << ii.first << " -> " << ii.second << ") ";
    f = 0;
  }
  return os << "]";
}
template <class T1, class T2>
ostream& operator<<(ostream& os, priority_queue<T1, T2> v) {
  bool f = 1;
  os << "[";
  while (!v.empty()) {
    auto x = v.top();
    v.pop();
    if (!f) os << ", ";
    f = 0;
    os << x;
  }
  return os << "]";
}
template <class T1, class T2>
void printarray(const T1& a, T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
  cout << endl;
}
void cio() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
}
const long long mod = 998244353;
long long N;
long long p[1000500];
int main() {
  scanf("%lld", &N);
  long long ans = 1;
  for (long long i = 1; i <= N; i++) {
    ans = (ans * i) % mod;
  }
  ans = (ans * N) % mod;
  long long tmp = 1;
  for (int i = N; i >= 2; i--) {
    tmp = tmp * i % mod;
    ans = (ans - tmp + mod) % mod;
  }
  printf("%lld\n", ans);
}
