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;
inline void getInt(int* p);
const long long mod = 998244353ll;
long long qpow(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return ans;
}
long long fac[1000500];
void build() {
  fac[0] = 1;
  for (int i = 1; i <= 1000005; i++) {
    fac[i] = (fac[i - 1] * i) % mod;
  }
}
long long n;
int main() {
  cin >> n;
  build();
  if (n == 1ll) {
    printf("1\n");
  } else {
    long long ans = fac[n];
    vector<long long> v;
    long long tmp;
    for (long long i = 1ll; i <= n - 1; i++) {
      long long k = fac[n] * qpow(fac[n - i], mod - 2) % mod;
      tmp = k * (fac[n - i] - 1) % mod;
      ans += tmp;
      ans = (ans + mod) % mod;
    }
    ans = (ans + mod) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}
inline void getInt(int* p) {
  char ch;
  do {
    ch = getchar();
  } while (ch == ' ' || ch == '\n');
  if (ch == '-') {
    *p = -(getchar() - '0');
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 - ch + '0';
    }
  } else {
    *p = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 + ch - '0';
    }
  }
}
