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 MOD = 998244353;
void yes() {
  cout << "Yes"
       << "\n";
}
void no() {
  cout << "No"
       << "\n";
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
struct FAIL {
  FAIL() {
    cout << "CHANGE!!!"
         << "\n";
  }
};
long long pow_two(long long a, long long n, long long mod) {
  if (n == 0) {
    return mod == 1 ? 0 : 1;
  }
  if (n % 2 == 1) {
    return a * pow_two(a, n - 1, mod) % mod;
  }
  long long t = pow_two(a, n / 2, mod);
  return t * t % mod;
}
class Inv {
 public:
  Inv(int mod) : mod(mod) {}
  int inverse(int value) { return pow_two(value, mod - 2, mod); }

 private:
  int mod;
};
class Comb {
 public:
  Comb(int max_value, int mod) : mod(mod), inv(mod) {
    precalc_facts(max_value, mod);
  }
  void precalc_facts(int n, int mod) {
    facts.resize(n);
    inv_facts.resize(n);
    facts[0] = 1;
    inv_facts[0] = 1;
    for (int i = (1); i < (facts.size()); ++i) {
      facts[i] = (long long)facts[i - 1] * i % mod;
      inv_facts[i] = inv.inverse(facts[i]);
    }
  }
  long long comb(int n, int k) {
    assert(0 <= n && n < facts.size());
    if (n < k) {
      return 0;
    }
    long long res = facts[n];
    res *= inv_facts[k];
    res %= mod;
    res *= inv_facts[n - k];
    res %= mod;
    return res;
  }
  long long fact(int n) {
    assert(0 <= n && n < facts.size());
    return facts[n];
  }

 private:
  int mod;
  Inv inv;
  vector<int> facts;
  vector<int> inv_facts;
};
int main() {
  int n;
  cin >> n;
  Comb comb(n + 1, MOD);
  long long res = comb.fact(n) * n % MOD;
  for (int i = (1); i < (n); ++i) {
    res += MOD - comb.fact(i) * comb.comb(n, n - i) % MOD;
    res %= MOD;
  }
  assert(0 <= res && res < MOD);
  cout << res;
}
