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 maxn = (int)(1e6 + 6);
const long long mod = (long long)(998244353);
pair<int, int> a[maxn];
int fact[maxn], inv_fact[maxn];
inline int mul(long long a, long long b) { return (a * b) % mod; }
inline int add(int a, int b) {
  int ret = a + b;
  if (ret >= mod) {
    ret -= mod;
  }
  return ret;
}
inline int sub(int a, int b) {
  int ret = a - b;
  if (ret < 0) {
    ret += mod;
  }
  return ret;
}
inline int poww(long long a, long long b) {
  int x = 1, y = a;
  while (b > 0) {
    if (b % 2) {
      x = mul(x, y);
    }
    y = mul(y, y);
    b /= 2;
  }
  return (int)(x % mod);
}
void build() {
  fact[0] = 1;
  for (int i = 1; i < maxn; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  inv_fact[maxn - 1] = poww(fact[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) {
    inv_fact[i] = mul(inv_fact[i + 1], i + 1);
  }
}
int C(int n, int r) {
  if (n < r || min(n, r) < 0) {
    return 0;
  }
  int val1 = fact[n], val2 = mul(inv_fact[r], inv_fact[n - r]);
  return mul(val1, val2);
}
int main() {
  ios_base::sync_with_stdio(0);
  build();
  build();
  int n, res = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int curr = mul(C(n, i), fact[i]);
    curr = mul(curr, sub(fact[n - i], 1));
    res = add(res, curr);
  }
  res = add(res, fact[n]);
  cout << res << endl;
  return 0;
}
