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>
T bigmod(T a, T b) {
  if (b == 0) return 1LL;
  if (b % 2LL == 0) {
    T temp = bigmod(a, b / 2LL) % 998244353;
    return (temp * temp) % 998244353;
  } else {
    return ((bigmod(a, b - 1LL) % 998244353) * (a % 998244353)) % 998244353;
  }
}
long long fact[1000002];
void doFactorial(long long n) {
  fact[0] = 1LL;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
  }
}
long long nPr(long long n, long long r) {
  long long inverse = bigmod(fact[n - r], 998244353 - 2LL);
  return (fact[n] * inverse) % 998244353;
}
int main() {
  long long n;
  scanf("%lld", &n);
  if (n <= 2) {
    cout << n << endl;
    return 0;
  }
  doFactorial(n);
  long long temp = 0;
  long long aa = ((n - 1LL) * 3LL) % 998244353;
  long long bb = ((n - 2LL) * 3LL) % 998244353;
  temp = (aa + bb) % 998244353;
  long long ans = (nPr(n, n - 3LL) * temp) % 998244353;
  long long subtract = 0;
  long long prev = 1LL;
  for (long long i = n; i > 3LL; i--) {
    long long temp = (prev * i) % 998244353;
    subtract = (subtract + temp) % 998244353;
    prev = temp;
  }
  ans = (ans - subtract + 998244353) % 998244353;
  cout << ans << endl;
  return 0;
}
