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;
int n;
const long long int MOD = 998244353;
class Combination {
  long long int ppow(long long int i, long long int j) {
    long long int res = 1LL;
    while (j) {
      if ((j & 1LL)) {
        res *= i;
        if (res >= MOD) {
          res %= MOD;
        }
      }
      j >>= 1;
      i *= i;
      if (i >= MOD) {
        i %= MOD;
      }
    }
    return res;
  }

 public:
  vector<long long int> k;
  vector<long long int> r;
  void resize(int N) {
    k.resize(N + 2);
    r.resize(N + 2);
    k[0] = 1;
    for (int i = 1; i < N + 2; i++) {
      k[i] = k[i - 1];
      k[i] *= i;
      if (k[i] >= MOD) k[i] %= MOD;
    }
    long long int al = k[k.size() - 1];
    long long int iv = ppow(k[k.size() - 1], MOD - 2);
    r[k.size() - 1] = iv;
    for (int i = (int)(r.size()) - 2; i >= 0; i--) {
      r[i] = r[i + 1] * (i + 1);
      if (r[i] >= MOD) {
        r[i] %= MOD;
      }
    }
  }
  long long int C(int a, int b) {
    if (a < b) return 0;
    long long int up = k[a];
    long long int dw = r[b] * r[a - b];
    dw %= MOD;
    up *= dw;
    up %= MOD;
    return up;
  }
  long long int H(int a, int b) { return C(a + b - 1, b); }
  long long int catalan_number(int n) {
    return (C(2 * n, n) + MOD - C(2 * n, n - 1)) % MOD;
  }
} c;
int main() {
  cin >> n;
  c.resize(2000000);
  long long int subt = 0;
  for (int i = 1; i < n; i++) {
    long long int way = c.C(n, i);
    way *= c.k[n - i];
    way %= MOD;
    subt += way;
    if (subt >= MOD) subt %= MOD;
  }
  long long int whole = c.k[n];
  whole *= n;
  whole %= MOD;
  whole += MOD;
  whole %= MOD;
  whole += MOD - subt;
  whole %= MOD;
  printf("%lld\n", whole);
  return 0;
}
