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 = 1 << 29;
inline int two(int n) { return 1 << n; }
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int g = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
  }
}
long long modpow(long long a, long long b, long long md) {
  long long res = 1;
  a %= md;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % md;
    a = a * a % md;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const int sz = 1e6 + 5;
long long n, f[sz], inv[sz];
void prec() {
  for (int i = 0; i < n + 1; i++) {
    if (!i)
      f[i] = 1;
    else
      f[i] = (f[i - 1] * i) % 998244353;
  }
  inv[n] = modpow(f[n], 998244353 - 2, 998244353);
  for (int i = n - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1);
    inv[i] %= 998244353;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ifstream in("input.txt");
  ofstream out("output.txt");
  cin >> n;
  prec();
  long long s = 1, tmp;
  for (int k = 1; k < n + 1; k++) {
    tmp = 0;
    tmp = (f[n] * inv[k]) % 998244353;
    tmp *= (f[k] - 1);
    tmp %= 998244353;
    s += tmp;
    s %= 998244353;
  }
  cout << s << endl;
  return 0;
}
