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;
long long A1, A2, B1, B2, C1, C2, n, m, b, c, x, e, h, res1, a1, a2, w, a3, y,
    z, a4, a5, a6, a7, a8, a9, a10, last, LL, timer, start, finish, flag, step,
    tests, kilk, bnum;
string s, s1;
long long t;
long long res;
long double l, r;
pair<long long, long long> mas[1007], mas1[1007];
set<long long> answer;
int main() {
  cin >> n;
  res = 1;
  for (int i = 1; i <= n; i++) {
    res *= i;
    res %= 998244353;
  }
  res1 += res;
  a1 = 1;
  for (int i = n; i >= 1; i--) {
    a1 *= i;
    a1 %= 998244353;
    res1 += (res + 998244353 - a1) % 998244353;
    res1 %= 998244353;
  }
  cout << res1 << endl;
}
