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 MOD = 998244353;
long long bpow(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long result = bpow(x, y / 2);
  return result * result % MOD * (y % 2 == 1 ? x : 1) % MOD;
}
int32_t main() {
  long long n;
  cin >> n;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; ++i) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  vector<long long> C(n + 1);
  C[0] = 1;
  for (long long i = 1; i <= n; ++i) {
    C[i] = fact[n] * bpow(fact[i], MOD - 2) % MOD * bpow(fact[n - i], MOD - 2) %
           MOD;
  }
  long long answer = (n * fact[n] % MOD - (n - 1) + MOD) % MOD;
  vector<long long> f(n + 2);
  for (long long i = 1; i <= n; ++i) {
    f[i] = C[i] * fact[n - i] % MOD;
  }
  f[n + 1] = 0;
  for (long long i = 1; i < n; ++i) {
    answer = (answer - (f[i] - f[i + 1] + MOD) % MOD * i % MOD + MOD) % MOD;
  }
  cout << answer << '\n';
  return 0;
}
