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 int N = 1e6 + 10;
const int mod = 998244353;
int fact[N];
int inv[N];
int mul(int a, int b) { return ((long long)(a % mod) * (b % mod)) % mod; }
int add(int a, int b) { return ((a % mod) + (b % mod)) % mod; }
int fastpow(int base, int power) {
  if (!power) return 1;
  int ret = fastpow(base, power / 2);
  ret = mul(ret, ret);
  if (power & 1) ret = mul(ret, base);
  return ret;
}
int npr(int n, int r) { return mul(fact[n], inv[n - r]); }
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  fact[0] = fact[1] = inv[0] = inv[1] = 1;
  for (int i = 1; i < N; i++) fact[i] = mul(fact[i - 1], i);
  inv[N - 1] = fastpow(fact[N - 1], mod - 2);
  for (int i = N - 2; i > 1; i--) inv[i] = mul((i + 1), inv[i + 1]);
  int n;
  cin >> n;
  int ans = n;
  int mult = n - 2;
  int r = 1;
  for (int i = 2; i <= n - 1; i++) {
    ans = add(ans, mul(i, mul(mult, npr(n, r))));
    r++;
    mult--;
  }
  cout << ans;
  return 0;
}
